Define the Wu manifold $W(n) = SU(n)/SO(n)$, the inclusion $SO \to SU$ given by thinking of $\Bbb C^n = \Bbb R^n \otimes \Bbb C$ (that is, including real matrices into complex matrices). Note that $W(1) = *$, $W(2) = S^2$, and $W(3)$ is what is usually called the Wu manifold.

There is a natural map $W(n) \to W(n+1)$, including $SU(n) \to SU(n+1)$ and then 'quotienting by more'; in fact, because $SO(n+1) \cap SU(n) = SO(n)$, it is injective.

A long exact sequence calculation shows that $\pi_1 W(n) = 0$ for all $n$, while $\pi_2 W(n) = \Bbb Z/2$ for $n \geq 3$ and $\Bbb Z$ for $n = 2$; the natural map $\pi_2 W(n) \to \pi_2 W(n+1)$ is a surjection for all $n \geq 2$ and an isomorphism for $n \geq 3$. In particular, we have that the map $H^2(W(n+1);\Bbb Z/2) \to H^2(W(n);\Bbb Z/2)$ is an isomorphism for all $n \geq 2$. To determine if $TW(n+1)$ is spinnable or not (that is, if $w_2$ is zero or not), it suffices to check that of its restriction to $W(n)$.

Because $TW(n+1)\big|_{W(n)} = TW(n) \oplus NW(n)$, the cruciual inductive step is calculating $w_2(NW(n))$.

Now notice that the map $W(n) \to W(n+1)$ factors through $SU(n+1)/SO(n) =: E(n+1).$ In fact, this fits into a fiber sequence $S^n \to E(n+1) \to W(n+1)$, and so the map $W(n) \to W(n+1)$ comes equipped with a section of this sphere bundle. Allow me to phrase the necessary result in unnecessary generality: given a manifold $M$, a fiber bundle $F \to E \to B$, and an embedding $M \xrightarrow{i_E} E$ that projects to an embedding $i_B: M \to B$, then if we write $T^F E$ for the fiberwise tangent bundle, $N(i_E) = N(i_B) \oplus i_E^* T^FE$. When the fiber bundle is furthermore the unit sphere bundle of a vector bundle $V$ over $B$, then $T^F E \oplus \Bbb R = \pi^*V$.

Before continuing, then, we identify the rank $n+1$ vector bundle $V$ giving rise to the sphere bundle $S^n \to SU(n+1)/SO(n) \to SU(n+1)/SO(n+1)$. If $G$ is a topological group with closed subgroup $SO(n)$, then we have a fibration sequence $G/SO(n) \to BSO(n) \to BG$; the first map classifies the $SO(n)$-bundle $G \to G/SO(n)$, and the associated sphere bundle is $G \times_{SO(n)} (SO(n)/SO(n-1)) = G/SO(n-1)$, as desired. In particular, the vector bundle $V$ corresponding to $W(n+1)$ is classified by the fiber inclusion $W(n+1) \to BSO(n+1)$. Running the long exact sequence of homotopy groups of a fibration, we see that this is an isomorphism on $\pi_2$ for all $n$, and in particular, $w_2(V) \neq 0$.

Lastly, the normal bundle to $W(n)$ in $E(n+1)$ is trivial: there is an $SU(n)$-equivariant trivialization of the normal bundle to $SU(n)$ in $SU(n+1)$ (using the usual push-around-using-group-structure-argument), which descends to a trivialization of the normal bundle of $SU(n)/SO(n)$ in $SU(n+1)/SO(n)$, as desired.

Plugging all this together, we get the formula $\text{triv} = NW(n) \oplus i_E^*\pi^*V = NW(n) \oplus i_B^* V$; and because the map $i_B^*: H^2(W(n+1);\Bbb Z/2) \to H^2(W(n);\Bbb Z/2)$ is an isomorphism for $n+1 \geq 3$, we see that $w_2(NW(n)) = w_2(i_B^*V) \neq 0$ for all $n \geq 2$. Because this is precisely the difference between $w_2 TW(n+1)$ and $w_2 TW(n)$, and $H^2(W;\Bbb Z/2) = \Bbb Z/2$, we see that whether or not $W(n)$ is spin alternates in $n$, for $n \geq 2$.

As $W(2) = S^2$, we see that $W(n)$ is spin for even $n$ and fails to be spin for odd $n$.

Lastly, because one may verify that $\beta: H^2(W(3);\Bbb Z/2) \to H^3(W(3);\Bbb Z/2)$ is nontrivial by hand using a calculation of $H^*(W(3))$ and the Bockstein long exact sequence, we see via naturality that $\beta$ is nontrivial on the degree 2 class of $W(n)$ for all $n \geq 3$; in particular, $W(2n+1)$ is not even spin^c for $2n+1 \geq 3$.