Answer
$I_y=\frac{b^3h}{6}$
Work Step by Step
We can find the required moment of inertia as follows:
$dA=ydx$
Given that $y=\frac{h}{b^3}x^3$
$\implies dA=\frac{h}{b^3}x^3 dx$
Now $I_y=\int x^2 dA$
$\implies I_y=\int_0^b x^2 \frac{h}{b^3}x^3 dx$
$\implies I_y=\int_0^b \frac{h}{b^3}x^5dx$
$\implies I_y=\frac{h}{b^3}\frac{x^6}{6}|_0^b$
$\implies I_y=\frac{b^3h}{6}$