From second branch, we can see the potential drop across the branch is $5+\left(1\times 1\right)=6 \mathrm{V}$.

Now for the first branch, we can write

$15-IR=6 ...\left(1\right)$ and from third branch, we can write

$I_1\times 3=6$

$\Rightarrow I_1=2 \mathrm{A}$

Now from Kirchoff's junction rule,

$I=I_1+1=3$

Now, from equation(1), we can write

$15-3R=6$

$\Rightarrow R=3\Omega$

All three branches are in parallel, therefore we can write equivalent resistance as:

$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{3}+\frac{1}{3}+1$

$\Rightarrow R_{\mathrm{eq}}=\frac{3}{5} \Omega =0.6 \Omega$

As all branches are in parallel with the capacitor branch, potential drop across all three branches will be the same. Therefore,

$E_{\mathrm{eq}}=6 \mathrm{V}$.

Now, current variation in the circuit due to charging will be $\frac{6}{\left({\displaystyle \frac{3}{5}+3}\right)}{e}^{\frac{-\left(t-t_0\right)}{CR}}$

At, $t=t_0+7.2 \mathrm{\mu s}$, we get

$i=\frac{6\times 5}{18}{e}^{-\frac{\left(7.2\times {10}^{-6}\right)}{\left(2\times {10}^{-6}\right)\times \left(3.6\right)}}$

$$\begin{gathered} =\frac{30}{18}\times e^{-1} \\ =\frac{30}{18}\times 0.36 \end{gathered}$$

$=0.6 \mathrm{A}$

At steady state, voltage across capacitor $=6 \mathrm{V}$.

$Q=6\times 2=12\mu \mathrm{C}$.