= Solution
Here $|G|=p^3$, $Z=G'=\{(x,y,c):x=y=0\}\simeq C_p$. Degree divisibility and $\sum d_i^2=p^3$ exclude degrees strictly between $1$ and $p$. There are $p^2$ linear characters, corresponding to $G/G'$. For trivial $\Theta$, induction is their direct sum. For each of the $p-1$ nontrivial central characters, induction is $p$ copies of one degree-$p$ irreducible $V_\Theta$. Thus the complete list is $p^2$ degree-one and $p-1$ degree-$p$ representations; their squared degrees sum to $p^3$.
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