Answer
$I_z=1.53Kgm^2$
Work Step by Step
We can determine the required moment of inertia as follows:
$I_z=\frac{3\rho}{10}\Sigma \frac{1}{3}\pi r_i^2 h_i r_i^2$
$\implies I_z=\frac{\rho \pi}{10}\Sigma h_i r_i^4$
We plug in the known values to obtain:
$I_z=\frac{200}{10}(1.6(0.4)4-0.6(0.4)^4-0.8(0.2)^4)$
This simplifies to:
$I_z=1.53Kgm^2$
