Answer
$$ \bar{x}=\frac{11}{3}, \ \ \bar{y}=\frac{14}{27}$$
$$I_{y} =432$$
Work Step by Step
Since
\begin{align*}
M&=\int_{0}^{1} \int_{0}^{6}(x+y+1) d x d y\\
&=\int_{0}^{1}(6 y+24) d y\\
&= (3y^2+24y)\bigg|_{0}^{1}=27\\
M_{x}&=\int_{0}^{1} \int_{0}^{6} y(x+y+1) d x d y\\
&=\int_{0}^{1} y(6 y+24) d y\\
&= 2y^3+ 12y^2\bigg|_{0}^{1}=14\\
M_{y}&=\int_{0}^{1} \int_{0}^{6} x(x+y+1) d x d y\\
&=\int_{0}^{1}(18 y+90) d y\\
&=(9 y^2+90y)\bigg|_{0}^{1} =99
\end{align*}
Then
$$ \bar{x}=\frac{11}{3}, \ \ \bar{y}=\frac{14}{27}$$
and
\begin{align*}
I_{y}&=\int_{0}^{1} \int_{0}^{6} x^{2}(x+y+1) d x d y\\
&=216 \int_{0}^{1}\left(\frac{y}{3}+\frac{11}{6}\right) d y\\
&=432
\end{align*}
