Answer
a) ${\bf 75.9}\;\rm mA$
b) ${\bf 0.5}\;\rm ms$
Work Step by Step
$$\color{blue}{\bf [a]}$$
When moving the switch to position 2, we have an $LC$ circuit with a fully charged capacitor at $t=0$ s.
Recalling that the dynamics of the charge on the capacitor in $LC$-circuits is given by
$$Q=Q_0\cos(\omega t)$$
so the current in this circuit is given by
$$I=-\dfrac{ dQ}{dt}=-\dfrac{d}{dt}Q_0\cos(\omega t)$$
$$I= Q_0\omega \sin(\omega t)\tag 1$$
where $Q_0 \omega =I_0$ where $I_0$ is the maximum current.
Hence,
$$I_0= Q_0 \omega $$
when $Q_0=(\Delta V_C)_{\rm max} C$, and $\omega=\sqrt{1/LC}$,
$$I_0=\dfrac{ (\Delta V_C)_{\rm max} C}{\sqrt{LC}} $$
where $ (\Delta V_C)_{\rm max}=\Delta V_{\rm battery}$ since the capacitor was fully charged when it was connected to the battery.
$$I_0=\dfrac{ V_{\rm battery} C}{\sqrt{LC}} $$
$$I_0=\dfrac{ V_{\rm battery} \sqrt{C}}{\sqrt{L }} $$
Plug the known;
$$I_0=\dfrac{ (12) \sqrt{(2\times 10^{-6})}}{\sqrt{(50\times 10^{-3})}} $$
$$I_0=\color{red}{\bf 75.9}\;\rm mA$$
$$\color{blue}{\bf [b]}$$
From (1), the current is first maximized when $\sin(\omega t)=1$, hence
$$\omega t=\dfrac{\pi }{2}$$
$$ t=\dfrac{\pi }{2\omega}=\dfrac{\pi \sqrt{LC} }{2 }$$
Plug the known;
$$ t =\dfrac{\pi }{2 }\sqrt{(50\times 10^{-3})(2\times 10^{-6})} $$
$$t=\color{red}{\bf 0.5}\;\rm ms$$
Note that the current is a sinusoidal function, so the current is maximized at $T/4$ starting from $t=0$ where $T$ is the period.
So we can obtain the same result by finding the period.
