I know it's a theorem that every exact sequence of abelian groups splits, but I'm guessing this problem is supposed to be solvable by smaller guns. Given an exact sequence of abelian groups $0 \to A \to B \to \mathbb{Z}^{n} \to 0$, we wish to show that $B \cong A \times \mathbb{Z}^{n}$.
I really know only the basics about short exact sequences and splitting homomorphisms, so I'm not sure how to proceed here without just applying the big theorem.
$B/A \cong \mathbf{Z}^n$ means that $B/A$ is generated freely by some elements $b_1 + A, \dots, b_n + A$. Use this to get a map $\mathbf{Z}^n \to B$.