You have only two choices. Since $[G:\mathcal{Z}(G)]=4$ we either have that $G/\mathcal{Z}(G)\cong \mathbb{Z}_4$ or $G/\mathcal{Z}(G)\cong \mathbb{Z}_2\oplus\mathbb{Z}_2$ since those are, up to isomorphism, the only groups of order four. But, $G/\mathcal{Z}(G)\ncong\mathbb{Z}_4$ for if $G/\mathcal{Z}(G)$ is cyclic then $G$ is abelian and so $\left|G/\mathcal{Z}(G)\right|=1$. It follows that $G/\mathcal{Z}(G)\cong \mathbb{Z}_2\oplus\mathbb{Z}_2\cong\mathcal{K}_4$
P.S. If you aren't sure why those are the only two groups you need merely note that any group of order $p^2$ where $p$ is a prime is abelian and the rest follows from the fundamental theorem of finitely generated abelian groups.