Answer
$$
I_{x y}=\frac{m}{12} a^2
$$
Work Step by Step
Mass of differential element is $d m=\rho d V=\rho h x d y=\rho h(a-y) d y$.
$$
m=\int_m d m=\rho h \int_0^a(a-y) d y=\frac{\rho a^2 h}{2}
$$
Now using the parallel axis theorem:
$$
\begin{aligned}
d I_{x y} & =\left(d I_{x^{\prime} y}\right)_G+d m x_G y_G \\
& =0+(\rho h x d y)\left(\frac{x}{2}\right)(y) \\
& =\frac{\rho h^2}{2} x^2 y d y \\
& =\frac{\rho h^2}{2}\left(y^3-2 a y^2+a^2 y\right) d y \\
I_{x y}= & \frac{\rho h}{2} \int_0^a\left(y^3-2 a y^2+a^2 y\right) d y \\
= & \frac{\rho a^4 h}{24}=\frac{1}{12}\left(\frac{\rho a^2 h}{2}\right) a^2=\frac{m}{12} a^2
\end{aligned}
$$
