A 12 V battery of negligible internal resistance is connected in series with a 2 kΩ resistor to a node. Between this node and the other battery…

Replace everything seen by the capacitor by its Thevenin equivalent. The open-circuit voltage across the $4\ \mathrm{k\Omega}$ resistor is $V_{th}=12\times\dfrac{4}{2+4}=8\ \mathrm{V}$, and the equivalent resistance is $R_{th}=2\parallel4=\dfrac43\ \mathrm{k\Omega}$. The time constant is $\tau=R_{th}C=\dfrac43\times10^3\times100\times10^{-6}=0.1333\ \mathrm{s}$. The capacitor voltage is $V_C=V_{th}\left(1-e^{-t/\tau}\right)$, so setting $V_C=6\ \mathrm{V}$ gives $1-e^{-t/\tau}=0.75$, i.e. $t=\tau\ln4=0.1333\times1.3863=0.18484\ \mathrm{s}$. This is $184.84\ \mathrm{ms}$. The final value is $8\ \mathrm{V}$, not $12\ \mathrm{V}$, so $6\ \mathrm{V}$ is reachable. Quick check: at $t=\tau\ln4$ the exponential term equals $1/4$, so $V_C=8(1-0.25)=6\ \mathrm{V}$. Ignoring the $4\ \mathrm{k\Omega}$ branch and using $\tau=RC=0.2\ \mathrm{s}$ with $V_\infty=12\ \mathrm{V}$ would give the wrong value of $138.63\ \mathrm{ms}$.