Answer
$$ \bar{z}=\frac{M_{x y}}{M}=\frac{5}{6},,\ \bar{x}=\bar{y}=0,$$
Work Step by Step
Since
\begin{align*}
M&=\int_{0}^{2 \pi} \int_{0}^{4} \int_{0}^{\sqrt{r}} d z r d r d \theta\\
&=\int_{0}^{2 \pi} \int_{0}^{4} r^{3 / 2} d r d \theta\\
&=\frac{64}{5} \int_{0}^{2 \pi} d \theta=\frac{128 \pi}{5}\\
M_{x y}&=\int_{0}^{2 \pi} \int_{0}^{4} \int_{0}^{\sqrt{r}}d z r d r d \theta\\
& =\frac{1}{2} \int_{0}^{2 \pi} \int_{0}^{4} r^{2} d r d \theta\\
&=\frac{32}{3} \int_{0}^{2 \pi} d \theta\\
&=\frac{64 \pi}{3}
\end{align*}
Then by symmetry
$$ \bar{z}=\frac{M_{x y}}{M}=\frac{5}{6},,\ \bar{x}=\bar{y}=0,$$
