Answer
$I_z=34.2Kg\cdot m^2$
Work Step by Step
We can determine the required moment of inertia as follows:
The mass of the big cone is given as
$m=\rho V$
$\implies m=200(\frac{1}{3}\pi)(1.333)(0.8)^2$
$\implies m=178.72Kg$
The mass of the red cone is
$m=\rho (-V)$
$\implies m=200(\frac{-1}{3})(0.333)(0.2)^2=-2.79Kg$
and the mass of the depression cone is given as
$m=\rho(-V)$
$\implies m=200(-\frac{1}{3}\pi)(0.6)(0.2)^2=-5.03Kg$
Now $I_z=\Sigma I_z^{\prime}$
We plug in the known values to obtain:
$I_z=\frac{3}{10}(178.72)(0.8)^2+\frac{3}{10}(-2.79)(0.2)^2+\frac{3}{10}(-5.03)(0.2)^2$
This simplifies to:
$I_z=34.2Kg\cdot m^2$
