Let $M$ be a $\mathbb Z_3$-orientable $n$-manifold. The goal is to show that $M$ is $\mathbb Z$-orientable.
Assume that $M$ is closed and connected. Suppose, for a contradiction, that $M$ is not $\mathbb Z$-orientable. Then $H_n(M; \mathbb Z) = 0$. Moreover, the torsion subgroup of $H_{n-1}(M; \mathbb Z)$ is $\mathbb Z_2$ (these two facts follow from Hatcher, Ch $3.3$).
By Universal Coefficient Theorem, $H_n(M; \mathbb Z_3) \cong (H_n(M; \mathbb Z) \otimes \mathbb Z_3) \oplus \mathrm{Tor}(H_{n-1}(M; \mathbb Z), \mathbb Z_3) = 0$, which is a contradiction since $H_n(M; \mathbb Z_3) \cong \mathbb Z_3$ as $M$ is a closed, connected, $\mathbb Z_3$-orientable n-manifold.
At the moment, I am not sure how to approach this problem without assuming $M$ to be closed.
Connectedness shouldn't be an issue since I can take a component of $M$, which is still $\mathbb Z_3$-orientable.
I would appreciate any help/hint.