Answer
${\bf 3.93 }\;\rm V$
Work Step by Step
We know that moving the coil away from the flux zone, to a zero flux zone, 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 $\Phi=AB=\pi r^2 B$
Noting that the coil will be exposed to zero flux after pulling it out of the flux region, so $\Phi_f=0$,
$$\Delta q=\dfrac{N}{R}\left| 0-\pi r^2 B \right|$$
$$\Delta q=\dfrac{N}{R}\left| -\pi r^2 B \right|\tag 2$$
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),
$$\Delta V_C=\dfrac{N\left| -\pi r^2 B \right|}{RC}$$
Plug the known;
$$\Delta V_C=\dfrac{(10)\left| -\pi (0.005)^2 (1\times 10^{-3}) \right|}{(0.20)(1\times 10^{-6})}$$
$$\Delta V_C=\color{red}{\bf 3.93 }\;\rm V$$