= Solution
For the <eccentric circular-hole disk inertia>, let $D=\pi\rho R^2$ be the <mass> of the complete disc before removing material, $h=\pi\rho r_o^2$ the missing <mass>, and $W=D-h$ the wheel's remaining <mass>. Denote the complete disc's centre by C and the axle/hole centre by H; $CH=r_o$. Assume the hole fits inside the disc, $0<r_o<R/2$. The <parallel axis theorem> gives the full disc's <moment of inertia> about H as $\tfrac12DR^2+Dr_o^2$. Subtract the absent little disc about its own centre H:
$$
\boxed{I_H=\frac12DR^2+Dr_o^2-\frac12hr_o^2=\frac{\pi\rho}{2}\left(R^4+2R^2r_o^2-r_o^4\right).}
$$
The axle is a support, not part of the rotating wheel's stated <mass>. Its frictionless bearing does not alter this <moment of inertia>. For fixed-axis <angular velocity> $\omega$,
$$
\boxed{T_H=\frac12I_H\omega^2=\frac{\pi\rho\omega^2}{4}\left(R^4+2R^2r_o^2-r_o^4\right).}
$$
The missing material must be subtracted with its actual displacement relative to the axis; treating the wheel as a concentric annulus would give the wrong result.
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