Integral Stiefel-Whitney classes and stable almost complex structure

Question: Let $$\xi$$ be a real vector bundle over a space $$B$$. Suppose $$\xi$$ admits stable almost complex structure, i.e., $$\xi\oplus \epsilon^k$$ admits almost complex structure for some $$k\geq 0$$, where $$\varepsilon$$ denotes a trivial real vector bundle. Is it true that every integral Stiefel–Whitney class $$W_j(\xi)\in H^j(B;\Bbb Z)$$ of $$\xi$$ vanishes?

My attempt: Consider a complex vector bundle $$\omega$$ over $$B$$ such that $$\omega_\Bbb R=\xi\oplus \varepsilon_k$$, where $$\omega_\Bbb R$$ denotes the realification of $$\omega$$. Now, it is known that (see Proposition 3.8. of Allen Hatcher's Vector bundles and K-theory) $$w_{2j}(\omega_\Bbb R)$$ is the image of $$c_j(\omega)$$ under the coefficient homomorphism $$\rho_{2j}\colon H^{2j}(B;\Bbb Z)\to H^{2j}(B;\Bbb Z_2)$$ and $$w_{2j+1}(\omega_\Bbb R)=0$$ for all $$j\geq 0$$. Therefore, $$w_{2j+1}(\xi)=w_{2j+1}(\xi\oplus \varepsilon_k)=w_{2j+1}(\omega_\Bbb R)=0$$ for all $$j\geq 0$$. Further, $$w_{2j}(\xi)=w_{2j}(\xi\oplus \varepsilon_k)=w_{2j}(\omega_\Bbb R)=\rho_{2j}\big(c_j(\omega)\big)$$ for all $$j\geq 1$$. Thus, $$\mathrm W_{2j}(\xi)= \beta_{2j-1}\big(w_{2j-1}(\xi)\big)= \beta_{2j-1}\big(0\big)=0$$ and $$\mathrm W_{2j+1}(\xi)= \beta_{2j}\big(w_{2j}(\xi)\big)= \beta_{2j}\rho_{2j}\big(c_j(\omega)\big)=0$$ as $$\beta_{2j}\rho_{2j}=0$$ for all $$j\geq 1$$, where $$\beta_{n-1}\colon H^{n-1}(B;\Bbb Z_2)\to H^n(B;\Bbb Z)$$ is the $$(n-1)$$-th Bockstein of $$0\to \Bbb Z\xrightarrow{\times 2}\Bbb Z\to \Bbb Z/2\Bbb Z\to 0$$ for each $$n\geq1$$.

Am I doing something wrong? The paper Obstructions to the existence of almost complex structures by Massey, W. S. says that the vanishing of the odd integral Stiefel-Whitney classes is a well-known necessary condition for the existence of almost complex structure.

Your argument is correct. Massey only mentions the vanishing of the odd integral Stiefel-Whitney classes because, for an orientable bundle $$\xi$$, we have $$W_{2j+1}(\xi) = 0 \implies W_{2j+2}(\xi) = 0$$.
To see this, note that $$\rho_{2j+1}(W_{2j+1}(\xi)) = \rho_{2j+1}\beta_{2j}(w_{2j}(\xi)) = \operatorname{Sq}^1(w_{2j}(\xi))$$ where $$\operatorname{Sq}^1 : H^{2j}(B; \mathbb{Z}_2) \to H^{2j+1}(B; \mathbb{Z}_2)$$ is the first Steenrod square. By Wu's formula, we have $$\operatorname{Sq}^1(w_{2j}(\xi)) = w_1(\xi)w_{2j}(\xi) + w_{2j+1}(\xi)$$, so $$\rho_{2j+1}(W_{2j+1}(\xi)) = w_{2j+1}(\xi)$$ if $$\xi$$ is orientable. Therefore, if $$W_{2j+1}(\xi) = 0$$, then $$w_{2j+1}(\xi) = 0$$, and hence $$W_{2j+2}(\xi) = 0$$ as you observed.