Answer
${\bf 12}\;\rm V$
Work Step by Step
We know that rotating the coil through the flux zone will change the area exposed to the flux and this will create an induced current.
This induced current is given by
$$I_{\rm coil}=\dfrac{\varepsilon_{\rm coil}}{R}\tag 1$$
So the induced emf is given by Faraday’s law,
$$\varepsilon_{\rm coil}=N\left| \dfrac{d\Phi}{dt}\right|$$
Plug into (1);
$$I_{\rm coil}=\dfrac{N}{R}\left| \dfrac{d\Phi}{dt}\right|$$
Recalling that $I=dq/dt$, the author put it as a hint.
$$\dfrac{dq}{dt}=\dfrac{N}{R}\left| \dfrac{d\Phi}{dt}\right|$$
Hence,
$$dq=\dfrac{N}{R}\left| d\Phi \right|$$
Replacing $d$ by $\Delta $;
$$\Delta q=\dfrac{N}{R}\left| \Delta\Phi \right|$$
$$dq=\dfrac{N}{R}\left| \Phi_f-\Phi_i \right|$$
where $\Phi=AB\cos\theta=\pi r^2 B\cos\theta$
$$\Delta q=\dfrac{N}{R}\left| \pi r^2 B\cos\theta_f-\pi r^2 B\cos\theta_i \right|$$
Initially, the angle between the area vector of the coil and the magnetic flux is $90^\circ+60^\circ=150^\circ$, and after rotating the coil $180^\circ$, the angle between them is then $30^\circ$, see the figures below.
$$\Delta q=\dfrac{\pi r^2 B N}{R}\left| \cos30^\circ - \cos150^\circ \right|\tag 2$$
We need to find the final voltage across the capacitor due to the induced current after pulling out the coil.
Recalling that the voltage across the capacitor is given by
$$\Delta V_C=\dfrac{\Delta q}{C}$$
Plug $\Delta q$ from (2),
$$\Delta V_C=\dfrac{\pi r^2 B N\left| \cos30^\circ - \cos150^\circ \right| }{RC}$$
Plug the known;
$$\Delta V_C=\dfrac{\pi (0.02)^2 (55\times 10^{-6})(200 ) \left| \cos30^\circ - \cos150^\circ \right|}{(2.0)(1\times 10^{-6})}=11.97\;\rm V$$
$$\Delta V_C=\color{red}{\bf 12}\;\rm V$$
