= Solution
Set $x=r\phi$ and $a=2\pi r$. The wrapped spatial form is
$$
\Psi_n(r\phi)=\sqrt{\frac2a}\sin\frac{n\phi}{2},\qquad
\frac{\partial^2\Psi_n}{\partial\phi^2}=-\frac{n^2}4\Psi_n.
$$
Hence the differential expression for the <quantum rotor> acts as
$$
\hat H_{\mathrm R}\Psi_n=\frac{n^2\hbar^2}{8I}\Psi_n.
$$
The box expression becomes $E_n=n^2\hbar^2/(8mr^2)$, so consistency gives
$$
\boxed{I=mr^2,}
$$
the <moment of inertia> of a point mass a distance $r$ from the axis. Equivalently, $\partial_x=r^{-1}\partial_\phi$ transforms the kinetic operator directly. Being an eigenfunction of this differential expression is not yet sufficient: the ring boundary conditions must also hold.
For a wavefunction normalized with the angular measure $\mathrm d\phi$, use $\Phi_n(\phi)=\sqrt r\,\Psi_n(r\phi)=\pi^{-1/2}\sin(n\phi/2)$. This change of normalization convention does not change the <energy eigenvalue>.
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