### Rigid Body

Let us choose axes $x,y,z$ along the disks axes. Then for one of them $I_{xx}\equiv I_0=\frac{1}{2}mr^2$, $I_{yy}=I_{zz}=I_0/2$ (other components). For the others one just has to change $x$ to $y$ or to $z$. Thus, the total tensor of inertia is diagonal $I_{xx}=I_{yy}=I_{zz}=2I_0=mr^2$, that is, $I_{ij}=mr^2 \delta_{ij}$. This means that the moment of inertia with respect to \textit{arbitrary} axis passing through the common center is $mr^2$.