Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the potential difference through an inductor is given by
$$\Delta V_L=-L\dfrac{dI}{dt}$$
where $I=I_0\sin (\omega t)$, so
$$\Delta V_L=-I_0L\dfrac{d }{dt}\sin (\omega t)$$
$$\boxed{\Delta V_L=-\omega I_0L \cos(\omega t)}$$
$$\color{blue}{\bf [b]}$$
Maximum voltage occurs when $\cos(\omega t)=1$, so
$$(\Delta V_L)_{\rm max}= \omega I_0L $$
where $\omega=2\pi f$;
$$(\Delta V_L)_{\rm max}= 2\pi fI_0L $$
Solving for $I_0$;
$$I_0=\dfrac{(\Delta V_L)_{\rm max}}{ 2\pi fL}$$
Plug the known;
$$I_0=\dfrac{(0.20)}{-2\pi (500\times 10^3)(50\times 10^{-6})}$$
$$I_0=\color{red}{\bf 1.27}\;\rm mA$$
