1. Resistance of each side of the square

The total resistance of the wire is $16\,\Omega$, and it is bent into a square loop. So each side has resistance $R=\frac{16}{4}=4\,\Omega.$

2. Battery connection across one side

Let the square vertices be $A,B,C,D$ in order, and the battery is connected across side $AB$. A capacitor is connected across diagonal $AC$.

Since the capacitor is in steady state, no current flows through the capacitor branch. We only need the potential difference between $A$ and $C$ in the resistive network.

Between $A$ and $B$, there are two parallel paths through the square wire:

• direct side $AB = 4\,\Omega$
• the other three sides in series: $A\to D\to C\to B = 4+4+4=12\,\Omega$

Hence the external resistance across battery terminals is $R_{\text{ext}}=4\parallel 12=\frac{4\cdot 12}{4+12}=3\,\Omega.$

Including internal resistance $1\,\Omega$, total circuit resistance is $R_{\text{total}}=3+1=4\,\Omega.$

So battery current is $I=\frac{9}{4}=2.25\,\text{A}.$

Therefore terminal voltage across the square is $V_{AB}=I\,R_{\text{ext}}=2.25\times 3=6.75\,\text{V}.$

3. Potential difference across diagonal $AC$

The $12\,\Omega$ path has current $I_{12}=\frac{V_{AB}}{12}=\frac{6.75}{12}=0.5625\,\text{A}.$

Along path $A\to D\to C\to B$, each side is $4\,\Omega$, so voltage drop across each side is $\Delta V = I_{12}\times 4 = 0.5625\times 4 = 2.25\,\text{V}.$

Thus potential drop from $A$ to $C$ is across two such sides: $V_{AC}=2.25+2.25=4.5\,\text{V}.$

So the capacitor gets charged to $4.5\,\text{V}$.

4. Energy stored in the capacitor

Using $U=\frac{1}{2}CV^2,$ with $C=4\,\mu F$ and $V=4.5\,\text{V}$, $U=\frac{1}{2}\cdot 4\,\mu F\cdot (4.5)^2$ $=2\cdot 20.25\,\mu J$ $=40.5\,\mu J.$

Given that energy is $\dfrac{x}{2}\,\mu J$, $\frac{x}{2}=40.5 \quad \Rightarrow \quad x=81.$

5. Comparison with stored answer

Our derived answer is $81$, which matches the stored correct answer.