Answer
a) ${\bf 6.28\times 10^{-7}}\;\rm s$
b) ${\bf 0.50}\;\rm mA$
Work Step by Step
$$\color{blue}{\bf [a]}$$
Recalling that the dynamics of the charge on the capacitor in $LC$-circuits is given by
$$Q=Q_0\cos(\omega t)$$
So the capacitor is fully discharged when $\cos(\omega t)=0$,
Hence,
$$\cos(\omega t)=0$$
Hence,
$$\omega t=\dfrac{\pi}{2}$$
where $\omega=\sqrt{1/LC}$
$$\dfrac{t}{\sqrt{LC}}=\dfrac{\pi}{2}$$
So that
$$t=\dfrac{\pi}{2}\sqrt{LC}$$
Plug the known;
$$t=\dfrac{\pi}{2}\sqrt{(20\times 10^{-3})(8\times 10^{-12})}\tag 1$$
$$t=\color{red}{\bf 6.28\times 10^{-7}}\;\rm s$$
$$\color{blue}{\bf [b]}$$
We know that the current through the inductor as the capacitor charge oscillates is given by
$$I=-\dfrac{-dQ}{dt}=-\dfrac{d}{dt}Q_0\cos(\omega t)$$
$$I= Q_0\omega \sin(\omega t)$$
where $Q_0 =\Delta V_C C$, and $\omega=\sqrt{1/LC}$,
$$I= \dfrac{ \Delta V_C C}{\sqrt{LC}}\sin\left(\dfrac{t}{\sqrt{LC}}\right)$$
where we need the current when the capacitor is fully discharged, so we need to plug $t$ from (1);
$$I= \dfrac{ \Delta V_C C}{\sqrt{LC}}\sin\left(\dfrac{\pi\sqrt{LC}}{2\sqrt{LC}}\right)$$
$$I= \dfrac{ \Delta V_C C}{\sqrt{LC}}\sin\left(\dfrac{\pi }{2 }\right)$$
$$I= \dfrac{ \Delta V_C C}{\sqrt{LC}}$$
Plug the known;
$$I= \dfrac{(25)(8\times 10^{-12})}{\sqrt{(20\times 10^{-3})(8\times 10^{-12})}}$$
$$I= \color{red}{\bf 0.50}\;\rm mA$$
