$e^{3z} = \sum_{n=0}^{\infty} \frac{(3z)^n}{n!} = \sum_{n=0}^{\infty} \frac{3^n z^n}{n!}$. So $z^2 e^{3z} = \sum_{n=0}^{\infty} \frac{3^n z^{n+2}}{n!}$. For $z^5$: $n + 2 = 5$, so $n = 3$. Coefficient: $\frac{3^3}{3!} = \frac{27}{6} = \frac{9}{2}$.