30@Ajoy - I'm afraid thats not the answer.
Anyway, I did the sum after a lot of tries.
Initially, let the angle between the area vector and magnetic field be 0.
Therefore at any time t,
\theta =\omega t
\phi =\int \vec{B}.\vec{dS}
\phi = BAcos\omega t = \frac{B\pi r^{2}}{2}cos\omega t
\xi = -\frac{d}{dt}(\phi )=\frac{B\pi r^{2}\omega}{2}sin\omega t
Now, total work done per period (T) of rotation,
W=\int_{0}^{T}{\frac{\xi ^{2}}{R}dt}
\begin{align*} W &=\int_{0}^{T}{\frac{B^{2}\pi^{2} r^{4}\omega ^{2}}{4R}\sin ^{2}\omega tdt} \\ &= \frac{B^{2}\pi ^{2}r^{4}\omega^{2}}{8R}\int_{0}^{T}{2\sin^{2}\omega tdt}\\ &= \frac{B^{2}\pi ^{2}r^{4}\omega^{2}}{8R}\int_{0}^{T}{\left( 1-\cos2\omega t\right)dt}\\ &= \left( \frac{B^{2}\pi ^{2}r^{4}\omega^{2}}{8R}\right)T \end{align*}
\text{Mean Power (P)}= \frac{W}{T}=\frac{\left( B \omega\pi r^{2}\right)^{2}}{8R}
