Answer
$I=2.7\times10^{-2}kg.m^2$
Work Step by Step
The moment of inertia of the stool is calculated by $$I=I_{disk}+3I_{rod}$$
The solid circular disk has $I_{disk}=\frac{mr^2}{2}=\frac{1}{2}(1.2\times0.16^2)=1.54\times10^{-2}kg.m^2$
We treat each rod as particle, with all of its mass being located at a distance $r=0.16m$ from the axis.
$I_{rod}=mr^2=0.15\times0.16^2=3.84\times10^{-3}kg.m^2$
Therefore, $$I=2.7\times10^{-2}kg.m^2$$
