At time t=0 , a battery of 10 ~V is connected across points A and B...

Voltage across the capacitors will increase from 0 to $10 \mathrm{~V}$ exponentially. The voltage at time $t$ will be given by $$\begin{aligned}& V=10\left(1-e^{-t / \tau} \mathrm{C}\right)\\ & \text{ Here } \tau_c=C_{\text{\text{ net } }} R_{\text{net }} \\ & =\left(1 \times10^6\right)\left(4 \times10^{-6}\right)=4 \mathrm{~s}\\ & \therefore \quad V=10\left(1-e^{-t / 4}\right) \\ & \end{aligned}$$Substituting $V=4$ volt we have, 4=10\left(1-e^{-t / 4}\right) Substituting $V=4$ volt we have, 4=10\left(1-e^{-t / 4}\right) or $e^{-t / 4}=0.6=\frac{3}{5}$ Taking $\log$ both sides we have, -$\frac{t}{4}=\ln$3-\ln 5 or $\quad t=4(\ln 5-\ln 3)=2 \mathrm{~s}. Hence, the answer is 2 .