Answer
a) ${\bf 6.93}\;\rm ms$
b) ${\bf 3.47}\;\rm ms$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the discharge of a capacitor through a resistor is given by
$$Q=Q_0e^{-t/\tau}\tag 1$$
So we need to find the time when $Q=\frac{1}{2}Q_0$
$$\frac{1}{2}\color{red}{\bf\not} Q_0=\color{red}{\bf\not} Q_0e^{-t/\tau}$$
Hence,
$$\ln \left( \frac{1}{2}\right)=\dfrac{-t}{\tau}$$
Thus,
$$t=-\tau \ln \left( \frac{1}{2}\right)$$
Plug the given;
$$t=-10\ln \left( \frac{1}{2}\right)$$
$$t=\color{red}{\bf 6.93}\;\rm ms$$
$$\color{blue}{\bf [b]}$$
when the energy stored to half its initial value, where$ U=Q^2/2C$ gives the energy in the capacitor.
So,
$$Q_0=\sqrt{2CU_0}$$
So, when $U_f=0.5U$
$$\dfrac{Q}{Q_0}=\sqrt{\dfrac{\color{red}{\bf\not} 2\color{red}{\bf\not} CU}{\color{red}{\bf\not} 2\color{red}{\bf\not} CU_0}}=\sqrt{\dfrac{ U}{ U_0}}$$
where $U=\frac{1}{2}U_0$, hence
$$\dfrac{Q}{Q_0}=\sqrt{\dfrac{1}{2}}$$
Thus,
$$Q=\dfrac{Q_0}{\sqrt{2}}$$
Plug into (1),
$$\dfrac{\color{red}{\bf\not} Q_0}{\sqrt{2}}=\color{red}{\bf\not} Q_0e^{-t/\tau}$$
Hence,
$$\ln \left( \frac{1}{\sqrt{2}}\right)=\dfrac{-t}{\tau}$$
Hence,
$$t=-\tau\ln \left( \frac{1}{\sqrt{2}}\right)=-10\ln \left( \frac{1}{\sqrt{2}}\right)$$
$$t=\color{red}{\bf 3.47}\;\rm ms$$
