#### Answer

$I_y=780in ^4$

#### Work Step by Step

We can find the required moment of inertia as follows:
$dA=(4-y)d_x$
Given that $y^2=x$
$\implies y=x^{0.5}$
$\implies dA=(4-x^{0.5})dx$
Now $I_y=\int x^2 dA$
$\implies I_y=\int_0^{16} x^2(4-x^{\frac{1}{2}})dx$
$\implies I_y=4\frac{x^3}{3}|_0^{16}-\frac{x^{\frac{7}{2}}}{\frac{7}{2}}|_0^{16}$
$\implies I_y=4(\frac{4096}{3})-\frac{2}{7}(16384)$
$\implies I_y=780in ^4$