Answer
a) ${\bf 80}\;\rm \mu C $
b) ${\bf 2.30\times 10^{-4}}\;\rm s$
Work Step by Step
$$\color{blue}{\bf [a]}$$
After a long time, the charge is fully charged and then there is no current in the right branch. This means that the current that passes through the left loop is given by
$$I=\dfrac{V_B}{R_{eq}}=\dfrac{100}{60+40}=\bf 1 \;\rm A$$
We can see that the voltage across the capacitor is equal to the voltage at the 40-$\Omega$ resistor since they are in parallel.
Thus,
$$\Delta V_C=V_{40\Omega}=40I=40=\dfrac{Q}{C}$$
Hence,
$$Q=40C $$
Plug the known;
$$Q=40(2\times 10^{-6})=\color{red}{\bf 80}\;\rm \mu C $$
$$\color{blue}{\bf [b]}$$
After opening the switch, the capacitor discharges through two resistors of 40 $\Omega$ and 10 $\Omega$ that are now in series.
We know that the discharge of a capacitor through a resistor is given by
$$Q=Q_0e^{-t/\tau}$$
where $\tau=RC$, and $Q=0.1Q_0$
$$0.1\color{red}{\bf\not}Q_0=\color{red}{\bf\not}Q_0e^{-t/R_{eq}C}$$
$$\ln(0.1)=\dfrac{-t}{R_{eq}C}$$
$$t=-\ln(0.1)R_{eq}C$$
Plug the known;
$$t=-\ln(0.1)(40+10)(2\times 10^{-6})$$
$$t=\color{red}{\bf 2.30\times 10^{-4}}\;\rm s$$
