Answer
a) ${\bf 0.925}\;\rm V$
b) ${\bf 0}\;\rm V$
Work Step by Step
We have here a loop and a changing flux due to the oscillation of the magnetic field, so we must have an induced current and an induced emf.
We know that the induced current is given by
$$I_{\rm loop}=\dfrac{\varepsilon_{\rm loop}}{R}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm solenoid}}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos\theta=AB $ since $\theta=0^\circ$, so
$$d\Phi =A dB=\pi r^2dB$$
Plug into (2),
$$\varepsilon_{\rm loop}=\pi r^2_{\rm loop}\left|\dfrac{dB }{dt}\right| $$
where $B=20\times 10^{-9}\sin\omega t$ where $\omega =2\pi f$,
$$\varepsilon_{\rm loop}=(20\times 10^{-9} )\pi r^2_{\rm loop} \left|\dfrac{d }{dt} \sin(2\pi f t) \right| $$
$$\varepsilon_{\rm loop}=(20\times 10^{-9} )\pi r^2_{\rm loop}(2\pi f)\left| \cos(2\pi f t) \right| $$
$$\varepsilon_{\rm loop}=(40\times 10^{-9} )\pi^2 r^2_{\rm loop} f\left| \cos(2\pi f t) \right| \tag 3$$
$$\color{blue}{\bf [a]}$$
The maximum induced emf is at $\cos(2\pi f t)=1$, plug that into (3),
$$\varepsilon_{\rm max}=(40\times 10^{-9} )\pi^2 r^2_{\rm loop} f $$
Plug the known;
$$\varepsilon_{\rm max}=(40\times 10^{-9} )\pi^2 (12.5\times 10^{-2})^2 (150\times 10^6) $$
$$\varepsilon_{\rm max}=\color{red}{\bf 0.925}\;\rm V$$
$$\color{blue}{\bf [b]}$$
The maximum induced emf is at $\cos(2\pi f t)=1$, but when the loop is turned 90$^\circ$ to be perpendicular to the oscillating electric field, then the angle between $\vec A$ and $\vec B$ is $90^\circ$ and hence, $d\Phi= AdB \cos 90^\circ=0$, and hence, there is no magnetic flux through the loop, so that there is no induced emf.
$$\varepsilon_{\rm max}=\color{red}{\bf 0}\;\rm V$$
