#### Answer

$I_y=0.144Kgm^2$

#### Work Step by Step

We can determine the required moment of inertia as follows:
The mass of the thin plate is given as
$m=\sigma A$
$\implies m=10(0.4)^2$
$\implies m=1.6Kg$
and the mass of thin disk is given as
$m=\sigma (-A)$
$\implies m=10(-\pi (0.1)^2)$
$\implies m=-0.314Kg$
Now $I_y=\Sigma (I^{\prime}_y+md^2_x)$
We plug in the known values to obtian:
$I_y=2(\frac{1.6(0.4)^2}{12})+1.6(0.2)^2+2(\frac{-0.314(0.1)^2}{4}-0.314(0.2)^2)$
This simplifies to:
$I_y=0.144Kgm^2$