= Solution
The <magnetic dipole moment> is $\boldsymbol\mu=IA\hat{\mathbf n}$ and its magnetic <torque> is $\boldsymbol\tau_{\rm mag}=\boldsymbol\mu\times\mathbf B$. Its component along the rotation axis opposes the imposed rotation. For positive $\omega$, the external driving <torque> needed to keep the <angular velocity> constant is
$$
\tau_{\rm drive}(t)=IAB\sin\omega t=\frac{B^2A^2\omega}{R}\sin^2\omega t.
$$
Averaging over a complete rotation, with $\langle\sin^2\omega t\rangle=1/2$, gives
$$
\boxed{\langle\tau_{\rm drive}\rangle=\frac{B^2\pi^2r^4\omega}{2R}}.
$$
As an independent sign and dimension check, $\omega\langle\tau_{\rm drive}\rangle=R\langle I^2\rangle$. Thus the mechanical input <power> equals the mean <Joule heating>; a passive loop cannot supply a sustained driving <torque>. This average assumes the same negligible-<self-inductance> approximation as the current expression.
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