Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We have a loop that consists of a rotating semicircle and stationary wires.
We have here a constant flux and the semicircle, that rotates, and this rotation makes the area that is exposed to the flux change.
Hence, we must have an induced current and an induced emf.
The other part of the loop, the lower part, has nothing to do with the induced emf since it is not moving and the flux does not change there.
We know that the induced current is given by
$$I_{\rm loop}=\dfrac{\varepsilon_{\rm loop }}{R_{\rm loop}}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm }}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos(2\pi f t) $
$$ \Phi = \frac{1}{2}\pi r^2 B\cos(2\pi f t)$$
Plug into (2),
$$\varepsilon_{\rm loop}= \left|\dfrac{d }{dt}\frac{1}{2}\pi r^2 B\cos(2\pi f t)\right| $$
$$\varepsilon_{\rm loop}=\frac{1}{2}\pi r^2 B \left|\dfrac{d }{dt}\cos(2\pi f t)\right| $$
$$\varepsilon_{\rm loop}=\frac{2\pi f}{2}\pi r^2 B \left|-\sin(2\pi f t)\right| $$
$$\varepsilon_{\rm loop}= \pi^2 r^2 f B \sin(2\pi f t) $$
Plug into (1),
$$I_{\rm loop}=\dfrac{ \pi^2 r^2 f B \sin(2\pi f t) }{R_{\rm loop}} $$
where $R_{\rm loop}=R_{\rm bulb}$,
$$I_{\rm loop}=\dfrac{ \pi^2 r^2 f B \sin(2\pi f t) }{R_{\rm bulb}} $$
Plug the known
$$I_{\rm loop}=\dfrac{ \pi^2 (0.05)^2 (0.20)f \sin(2\pi f t) }{(1)} $$
$$\boxed{I_{\rm loop}= (5\times 10^{-3}) \pi^2 f \sin(2\pi f t) } $$
$$\color{blue}{\bf [b]}$$
The maximum induced current occurs when $\sin(2\pi ft)=1$, so
$$I_{\rm max}= (5\times 10^{-3}) \pi^2 f $$
Hence,
$$f=\dfrac{I_{\rm max}}{ (5\times 10^{-3}) \pi^2 }\tag 3$$
We know that the power dissipated in the bulb is given by
$$P=I^2R_{\rm bulb}$$
To have the full light of the bulb,
$$P=I_{\rm max}^2R_{\rm bulb}$$
Thus,
$$ I_{\rm max}=\sqrt{\dfrac{P}{R_{\rm bulb}}}$$
Plug into (3),
$$f=\dfrac{\sqrt{\dfrac{P}{R_{\rm bulb}}}}{ (5\times 10^{-3}) \pi^2 } $$
Plug the known;
$$f=\dfrac{\sqrt{\dfrac{4}{1}}}{ (5\times 10^{-3}) \pi^2 }$$
$$f=\color{red}{\bf 405}\;\rm Hz $$
