It is known that if $G$ is a nonabelian $p$-group of order $p^n$, with an abelian subgroup of index $p$, then the number $k(G)$ of conjugacy classes of $G$ can be as large as $p^{n-1} + p^{n-2} - p^{n-3}$, with equality if and only if $|G'| = p$, where $G'$ is the commutator subgroup of $G$. For $n = 5$, we have $k(G) \leq p^4 + p^3 - p^2$, and this upper limit is reached if and only if $|G'| = p$.

My question is, if for the case $n = 5$, $|G'| = p$, how large can the centre $Z(G)$ be? Of course, $|Z(G)| \leq p^3$, but we can get any better information?