Answer
$7.44\times10^{-3}\cos(120\pi t)$
Work Step by Step
We have here a coil and a changing flux due to the change in the current of the solenoid, so we must have an induced current and an induced emf.
We know that the induced current is given by
$$I_{\rm coil}=\dfrac{\varepsilon_{\rm coil}}{R}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm coil}=\left|\dfrac{d\Phi_{\rm solenoid}}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos\theta=AB $ since $\theta=0^\circ$, so for
$$d\Phi =A_{\rm solenoid } dB_{\rm solenoid}$$
We used the area of the solenoid not that of the loop since we assumed that the solenoid is an ideal one and the flux outside its area is zero.
where $B_{\rm solenoid}=\mu_0 N_{\rm solenoid}I_{\rm solenoid}/l$
$$d\Phi_{\rm solenoid} =A_{\rm solenoid}\dfrac{\mu_0N_{\rm solenoid}dI_{\rm solenoid}}{l}$$
and for $N$ turns of the coil.
$$d\Phi_{\rm solenoid} =\pi N_{\rm coil} r^2_{\rm solenoid}\dfrac{\mu_0N_{\rm solenoid}dI_{\rm solenoid}}{l}$$
Plug into (2),
$$\varepsilon_{\rm coil}=\dfrac{\pi N_{\rm coil} r^2_{\rm solenoid}\mu_0N_{\rm solenoid}}{l}\left|\dfrac{dI_{\rm solenoid}}{dt}\right| $$
where $I_{\rm solenoid }=0.5\sin(2\pi f t)$
$$\varepsilon_{\rm coil}=\dfrac{\pi N_{\rm coil} r^2_{\rm solenoid}\mu_0N_{\rm solenoid}}{l}\left|\dfrac{d }{dt}0.5\sin(2\pi f t)\right| $$
$$\varepsilon_{\rm coil}=\dfrac{\pi N_{\rm coil} r^2_{\rm solenoid}\mu_0N_{\rm solenoid}}{l}\left| 0.5(2\pi f)\cos(2\pi f t)\right| $$
$$\varepsilon_{\rm coil}=\dfrac{\pi^2 f N_{\rm coil} r^2_{\rm solenoid}\mu_0N_{\rm solenoid}}{l}\left| \cos(2\pi f t)\right| $$
Plug into (1),
$$I_{\rm coil}=\dfrac{\dfrac{\pi^2 f N_{\rm coil} r^2_{\rm solenoid}\mu_0N_{\rm solenoid}}{l}\left| \cos(2\pi f t)\right| }{R} $$
Plug the known;
$$I_{\rm coil}=\dfrac{\dfrac{\pi^2 (60)(50)(0.01)^2(4\pi\times 10^{-7})(200)}{0.20} \cos(2\pi \times 60 t) }{0.50} $$
$$\boxed{I_{\rm coil}=7.44\times10^{-3}\cos(120\pi t)}$$
