Answer
\begin{align*}
I_{x} &=208 \\
I_{y}&= 280\\
I_{z}&= 360
\end{align*}
Work Step by Step
The plane $z=\frac{4-2 y}{3}$ is the top of the wedge, so
\begin{align*}
I_{x}&=\int_{-2}^{3} \int_{-2}^{4} \int_{-2}^{4} \int_{-4 / 3}^{(4-2 y) / 3}\left(y^{2}+z^{2}\right) d z d y d x \\
&=\int_{-3}^{3} \int_{-2}^{4}\left[\frac{8 y^{2}}{3}-\frac{2 y^{3}}{3}+\frac{8(2-y)^{3}}{81}+\frac{64}{81}\right] d y d x\\
&=\int_{-3}^{3} \frac{104}{3} d x=208 \\
I_{y}&=\int_{-3}^{3} \int_{-2}^{4} \int_{-4 / 3}^{(4-2 y) / 3}\left(x^{2}+z^{2}\right) d z d y d x\\
&=\int_{-3}^{3} \int_{-2}^{4}\left[\frac{(4-2 y)^{3}}{81}+\frac{x^{2}(4-2 y)}{3}+\frac{4 x^{2}}{3}+\frac{64}{81}\right] d y d x\\
&=\int_{-3}^{3}\left(12 x^{2}+\frac{32}{3}\right) d x\\
&=280\\
I_{z}&=\int_{-3}^{3} \int_{-2}^{4} \int_{-4 / 3}^{(4-2 y) / 3}\left(x^{2}+y^{2}\right) d z d y d x\\
&=\int_{-3}^{3} \int_{-2}^{4}\left(x^{2}+y^{2}\right)\left(\frac{8}{3}-\frac{2 y}{3}\right) d y d x\\
&=12 \int_{-3}^{3}\left(x^{2}+2\right) d x\\
&=360
\end{align*}
