Answer
$$I_x=\frac{M}{3}\left(b^{2}+c^{2}\right),\ \ I_{y}=\frac{M}{3}\left(a^{2}+c^{2}\right),\ \ I_{z}=\frac{M}{3}\left(a^{2}+b^{2}\right),$$
Work Step by Step
Since
\begin{align*}
I_{x}&=\int_{0}^{a} \int_{0}^{b} \int_{0}^{c}\left(y^{2}+z^{2}\right) d z d y d x\\
&=\int_{0}^{a} \int_{0}^{b}\left(c y^{2}+\frac{c^{3}}{3}\right) d y d x\\
&=\int_{0}^{a}\left(\frac{c b^{3}}{3}+\frac{c^{3} b}{3}\right) d x\\
&=\frac{a b c\left(b^{2}+c^{2}\right)}{3}\\
&=\frac{M}{3}\left(b^{2}+c^{2}\right)
\end{align*}
where $M=a b c ; $ and, by symmetry $$I_{y}=\frac{M}{3}\left(a^{2}+c^{2}\right),\ \ I_{z}=\frac{M}{3}\left(a^{2}+b^{2}\right)$$
