Before the key is closed ($t<t_0$). The capacitor branch is disconnected, so only the top two loops carry current. Let $V$ be the potential of the common node (relative to the other rail). Through the $1\Omega$ branch the given current is $1$ A flowing in a fixed direction, so $V=5+1\cdot 1=6$ V.

For the $3\Omega$ branch in series with the lower rail, the current $I_1$ satisfies $V-3I_1=0\Rightarrow I_1=\frac{6}{3}=2$ A. Option (B) holds.

Resistance $R$. KCL at the node: total current into the node from the $15$ V branch equals the sum of currents leaving through the other two branches: $I=1+2=3$ A. Then $15-IR=V=6$, giving $IR=9$ and $R=3\Omega$. Option (A) holds.

After the key is closed. Replace the circuit external to the capacitor with its Thevenin equivalent at the capacitor terminals. The three sources combine to give ${\epsilon }_{eq}=\frac{15/3+5/1+0/3}{1/3+1/1+1/3}=\frac{10}{5/3}=6$ V. The Thevenin resistance is ${\left(\frac{1}{3}+\frac{1}{1}+\frac{1}{3}\right)}^{-1}=\frac{3}{5}=0.6\Omega$. The capacitor sits in series with the additional $3\Omega$ in its branch, so the total series resistance is $3+0.6=3.6\Omega$.

Steady-state charge. $Q_{\infty }=C{\epsilon }_{eq}=2\mu \text{F}\times 6\text{V}=12\mu$C. Option (D) holds.

Time constant and current. $\tau =RC=3.6\times 2=7.2\mu$s. The charging current is $i(t)=\frac{{\epsilon }_{eq}}{R_{eff}}{e}^{-t/\tau }=\frac{6}{3.6}{e}^{-t/\tau }$. At $t=t_0+7.2\mu$s, $t/\tau =1$, so $i=\frac{6}{3.6}\times 0.36\approx 0.6$ A. Option (C) holds.

Correct options: (A), (B), (C), (D).