Answer
See the detailed answer below.
Work Step by Step
Initially, the circuit is only in the left loop which consists of the battery and the capacitor. And since the switch is at a for a long time, the capacitor is fully charged and its charge is given by
$$\Delta V_C=\dfrac{Q}{C}$$
where $\Delta V_C=\Delta V_{\rm battery}$
$$\Delta V_{\rm battery}=\dfrac{Q}{C}$$
$$Q_0=\Delta V_{\rm battery}C\tag 1$$
This is also the charge in the capacitor just Immediately after the switch is moved to the b.
$$\color{blue}{\bf [a]}$$
Plug the known into (1);
$$Q_0=(9)(4\times 10^{-6})=\color{red}{\bf 36}\;\rm \mu C$$
The initial current is then given by
$$I_0=\dfrac{\Delta V_R}{R}$$
where the capacitor is connected in parallel to the resistor. Just Immediately after the switch is moved to the b, the voltage of the capacitor is still equal to that of the battery. Hence the resistor must have the same initial voltage.
Hence,
$$I_0=\dfrac{9}{25}=\color{red}{\bf 0.36}\;\rm A$$
$$\color{blue}{\bf [b]}$$
The decaying of the charge in the capacitor occurs after the switch is moved to b.
This decay is given by
$$Q=Q_0e^{-t/\tau}$$
where $\tau=RC$
$$Q=Q_0e^{-t/RC}\tag 2$$
Plug the known;
$$Q=(36\mu)e^{-50\mu/(25)(4\mu)}$$
$$Q=\color{red}{\bf 21.8}\;\rm \mu C$$
And the current is given by
$$I=I_0e^{-t/RC}\tag 3$$
Plug the known;
$$I=(0.36)e^{-50\mu/(25)(4\mu)}\tag 3$$
$$I=\color{red}{\bf 0.22}\;\rm A$$
$$\color{blue}{\bf [c]}$$
We just need to use (2) and (3) again with changing the time only.
$$Q=Q_0e^{-t/RC} =(36\mu)e^{-200\mu/(25)(4\mu)}$$
$$Q=\color{red}{\bf 4.87}\;\rm \mu C$$
$$I=I_0e^{-t/RC} =(0.36)e^{-200\mu/(25)(4\mu)}\tag 3$$
$$I=\color{red}{\bf 49}\;\rm mA$$
