A battery of emf 12 V and negligible internal resistance is connected in series with a resistor R1=2 kΩ and a parallel combination of a resistor R2=4…

At $t=0$ the uncharged capacitor behaves like a wire across $R_2$, so $I_0=12/R_1=6$ mA, and (A) is true. In the steady state no current enters the capacitor; the current is $12/(2+4)\ \mathrm{k\Omega}=2$ mA and the capacitor voltage equals the drop across $R_2$, $2\ \mathrm{mA}\times4\ \mathrm{k\Omega}=8$ V, so $Q=CV=80\ \mu\mathrm{C}$ and (B) is true. The time constant uses the Thevenin resistance seen by the capacitor, $R_1\parallel R_2=\frac43\ \mathrm{k\Omega}$, giving $\tau=13.3$ ms, not $60$ ms (C is false). Final energy is $\frac12CV^2=\frac12\times10\times10^{-6}\times64=0.32$ mJ, and $0.72$ mJ would correspond to the full $12$ V (D is false).