Let $M$ be a manifold. The total StiefelWhitney class of $M$ is defined to be the StiefelWhitney class of the tangent bundle $TM$ $$ w(M)=1+w_1(TM)+w_2(TM)+\cdots $$ I want to find references for $$ w(SO(n)/SO(k)),(i.e. w(V_{nk}(\mathbb{R}^n))), \\ w(U(n)/U(k)),(i.e. w(V_{nk}(\mathbb{C}^n))), \\ w(Sp(n)/Sp(k)),(i.e. w(V_{nk}(\mathbb{H}^n))),\\ w(SO(n)/(SO(k)\times SO(nk))),(i.e. w(G_{nk}(\mathbb{R}^n))), \\ w(U(n)/(U(k)\times U(nk))),(i.e. w(G_{nk}(\mathbb{C}^n))), \\ w(Sp(n)/(Sp(k)\times Sp(nk))),(i.e. w(G_{nk}(\mathbb{H}^n))). $$ Which ones of the above are known? Where could I find these formulas?

2$\begingroup$ Have you looked in any of the standard characteristic classes references, such as Milnor and Stasheff? My impression is most of your questions are answered, or at least close to answered, there. $\endgroup$ – Ryan Budney Oct 12 '15 at 2:44

$\begingroup$ @RyanBudney Yes, Prof. Ryan. I found that $w$ is expressed in terms of $Sq$'s and Wu class $v$ [Th. 11.14]. However, I do not know how to continue... $\endgroup$ – QSR Oct 12 '15 at 3:43

$\begingroup$ To continue with the Wu formula you need to understand the mod 2 cohomology rings together with Poincaré duality and the action of the Steenrod algebra. This should certainly be tractable in your last 2 examples (and maybe your last 3, did you really mean to ask about the oriented Grassmannian there)? I agree this should be worked out somewhere. $\endgroup$ – Mark Grant Oct 12 '15 at 7:21

$\begingroup$ Aren't they in BorelHirzebruch, characteristic classes and homogeneous spaces? $\endgroup$ – user43326 Oct 12 '15 at 9:58

1$\begingroup$ @MarkGrant Actually I meant part I, chapter V. Of course, they dealt with only the case of $G_{n1}(F^n)$, and depending on $F$, they calculate Chern and Pontryagin classes, but one should be able to deduce the StiefelWhitney classes from them (for example, by taking "square root" of Chern classes). $\endgroup$ – user43326 Oct 12 '15 at 13:21
The Stiefel manifolds are all parallelizable for $nk\ge2$, so their total StiefelWhitney classes are equal to $1$. A reference is Theorem 3.1 of this paper by Kee Yuen Lam.
For the finite Grassmannians, things are a little more complicated. In the complex and symplectic cases you should be able to calculate these using the Wu formula. Partial results on the normal StiefelWhitney classes $\bar{w}(G_{nk}(\mathbb{R}^n))$ are scattered throughout the literature, a recent reference being
Korba?, J.; Novotny, P. On the dual StiefelWhitney classes of some Grassmann manifolds. Acta Math. Hungar. 123 (2009), no. 4, 319–330.
This article and the articles it references are content with showing nontriviality of some normal SWclass in order to deduce nonimmersion results, and do not give formulae for $w(G_{nk}(\mathbb{R}^n))$.
Most of the methods seem to use the vector bundle isomorphisms \begin{align*} T(G_{nk}(\mathbb{R}^n)) & \cong \operatorname{Hom}(\gamma_k,\gamma_{nk})\\ & \cong \gamma_k^*\otimes \gamma_{nk} \\ & \cong \gamma_k\otimes \gamma_{nk}\end{align*} and the splitting principle.
In this note, the first two StiefelWhitney classes of unoriented, oriented, and complex grassmannians are determined in terms of the StiefelWhitney classes of their tautological bundles. This is achieved via the method mentioned at the end of Mark Grant's answer.
For the unoriented grassmannian $\operatorname{Gr}(m, m+n) = O(m+n)/(O(m)\times O(n))$, we have
\begin{align*} w_1(\operatorname{Gr}(m, m+n)) &= (m + n)w_1(\gamma)\\ w_2(\operatorname{Gr}(m, m + n)) &= \left[\binom{m}{2} + \binom{n}{2} + m^2 + mn  1\right]w_1(\gamma)^2 + (m^2 + n^2)w_2(\gamma)\\ &= \begin{cases} 0 & m  n \equiv 2 \bmod 4\\ w_2(\gamma) & m  n \equiv 1 \bmod 4\\ w_1(\gamma)^2 & m  n \equiv 0 \bmod 4\\ w_2(\gamma) + w_1(\gamma)^2 & m  n \equiv 3 \bmod 4. \end{cases} \end{align*}
For the oriented grassmannian $\operatorname{Gr}^+(m, m+n) = SO(m+n)/(SO(m)\times SO(n))$, we have
\begin{align*} w_1(\operatorname{Gr}^+(m, m + n)) &= 0\\ w_2(\operatorname{Gr}^+(m, m + n)) &= \begin{cases} 0 & m  n \equiv 0 \bmod 2\\ w_2(\gamma_+) & m  n \equiv 1 \bmod 2. \end{cases} \end{align*}
For the complex grassmannian $\operatorname{Gr}^{\mathbb{C}}(m, m + n) = U(m + n)/(U(m)\times U(n))$, we have
\begin{align*} w_1(\operatorname{Gr}^{\mathbb{C}}(m, m + n)) &= 0\\ w_2(\operatorname{Gr}^{\mathbb{C}}(m, m + n)) &= (m + n)w_2(\gamma_{\mathbb{C}}). \end{align*}
As for the quaternionic grassmanian $\operatorname{Gr}^{\mathbb{H}}(m, m + n) = Sp(m + n)/(Sp(m)\times Sp(n))$, it is $2$connected, so $w_1(\operatorname{Gr}^{\mathbb{H}}(m, m + n)) = 0$ and $w_2(\operatorname{Gr}^{\mathbb{H}}(m, m + n)) = 0$.
In principle, you can calculate all of the StiefelWhitney classes using the splitting principle argument, but the calculations get more and more complicated. A short cut for the third StiefelWhitney class is to use the fact that $\operatorname{Sq}^1(w_2) = w_3 + w_1w_2$. It follows that we have
\begin{align*} w_3(\operatorname{Gr}(m, m + n)) &= \begin{cases} 0 & m  n \equiv 0, 2 \bmod 4\\ w_3(\gamma) & m  n \equiv 1 \bmod 4\\ w_3(\gamma) + w_1(\gamma)^3 & m  n \equiv 3 \bmod 4 \end{cases}\\ & \\ w_3(\operatorname{Gr}^+(m, m + n)) &= \begin{cases} 0 & m  n \equiv 0 \bmod 2\\ w_3(\gamma_+) & m  n \equiv 1 \bmod 2 \end{cases}\\ & \\ w_3(\operatorname{Gr}^{\mathbb{C}}) &= 0\\ & \\ w_3(\operatorname{Gr}^{\mathbb{H}}) &= 0. \end{align*}