A 12 V battery of negligible internal resistance is connected in series with a 2.0 kΩ resistor. Across the far end of this resistor and the battery's…

Replace everything seen by the capacitor with its Thevenin equivalent. The open-circuit voltage is the divider output $V_{th}=12\times\dfrac{3}{2+3}=7.2\ \mathrm{V}$ and the resistance is the parallel combination $R_{th}=\dfrac{2\times3}{2+3}=1.2\ \mathrm{k\Omega}$. The capacitor therefore charges towards $7.2\ \mathrm{V}$, not $12\ \mathrm{V}$, with time constant $\tau=R_{th}C=1200\times50\times10^{-6}=60\ \mathrm{ms}$: $V_C=V_{th}\left(1-e^{-t/\tau}\right)$. Setting $V_C=4.8\ \mathrm{V}$ gives $1-e^{-t/\tau}=\tfrac{4.8}{7.2}=\tfrac23$, so $e^{-t/\tau}=\tfrac13$ and $t=\tau\ln3=60\times1.0986=65.92\ \mathrm{ms}$. A direct numerical integration of the original circuit equations, $C\,dV/dt=(12-V)/R_1-V/R_2$, confirms this result. Note that the steady current through the $3.0\ \mathrm{k\Omega}$ resistor limits the final capacitor voltage to $7.2\ \mathrm{V}$, so a naive $RC$ with $2.0\ \mathrm{k\Omega}$ and $12\ \mathrm{V}$ would give the wrong time.