Four capacitors and a switch S are connected to a source as shown in the figure. Initially, S is open and the capacitors are uncharged after S is closed and steady state is reached, the energy stored in the 4mF capacitor in the units of ${10}^{-5}$J is

From the diagram 10 $\mu F$ and $10\mu F$ are series , Equivalent= 5 $\mu F$

$5\mu F and 5\mu F are in parallel$ .Equivalent is 10 $\mu F$

Now circuit has effectively two capacitors 4 $\mu$F and 10 $\mu F$ connected in series with a 14V  battery

Potential across 4$\mu F$ is $\frac{14}{10+4}\left(10\right)=10V$

energy= $\frac{1}{2}4\times {10}^{-6}\times 100= 20\times {10}^{-5}$