Answer
$$ \bar{x}=\frac{100}{9},\ \ \bar{y}=0,\ \ I_x= 20$$
Work Step by Step
Since
\begin{align*}
M&=\int_{0}^{20} \int_{-1}^{1}\left(1+\frac{x}{20}\right) d y d x\\
&=\int_{0}^{20}\left(2+\frac{x}{10}\right) d x\\
&=60 \\
M_{x}&=\int_{0}^{20} \int_{-1}^{1} y\left(1+\frac{x}{20}\right) d y d x\\
&=\int_{0}^{20}\left[\left(1+\frac{x}{20}\right)\left(\frac{y^{2}}{2}\right)\right]_{-1}^{1} d x\\
&=0\\
M_{y}&=\int_{0}^{20} \int_{-1}^{1} x\left(1+\frac{x}{20}\right) d y d x\\
&=\int_{0}^{20}\left(2 x+\frac{x^{2}}{10}\right) d x\\
&=\frac{2000}{3}
\end{align*}
Then
$$ \bar{x}=\frac{100}{9},\ \ \bar{y}=0$$
and
\begin{align*}
I_{x}&=\int_{0}^{20} \int_{-1}^{1} y^{2}\left(1+\frac{x}{20}\right) d y d x\\
&=\frac{2}{3} \int_{0}^{20}\left(1+\frac{x}{20}\right) d x=20
\end{align*}
