1. Find resistance of each side of the square

The total resistance of the wire is $16\,\Omega$ and it is bent into a square of 4 equal sides.

So, resistance of each side is $R=\frac{16}{4}=4\,\Omega$

2. Identify the terminals of the battery

The 9 V battery with internal resistance $1\,\Omega$ is connected across one side of the square, say side $AB$.

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

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

Thus, the square loop between $A$ and $B$ is equivalent to $R_{AB}=4\parallel 12=\frac{4\cdot 12}{4+12}=\frac{48}{16}=3\,\Omega$

3. Include internal resistance and find current from the battery

The battery sees external resistance $3\,\Omega$ in series with internal resistance $1\,\Omega$.

So total resistance is $R_{\text{total}}=3+1=4\,\Omega$

Hence current supplied by the battery is $I=\frac{9}{4}=2.25\,\text{A}$

4. Find terminal voltage across the square

The potential difference across the square loop (between $A$ and $B$) is $V_{AB}=I\times 3=2.25\times 3=6.75\,\text{V}$

(Equivalently, $V_{AB}=9-Ir=9-2.25\times 1=6.75\,\text{V}$.)

5. Find current in each branch of the square

Since $6.75$ V is across both branches:

• Current through side $AB$ ($4\,\Omega$ branch): $I_1=\frac{6.75}{4}=1.6875\,\text{A}$

• Current through the other three sides ($12\,\Omega$ branch): $I_2=\frac{6.75}{12}=0.5625\,\text{A}$

6. Find potentials at the opposite corners

Let the corners be $A,B,C,D$ in order, with battery connected across side $AB$.

Take $V_B=0$ and $V_A=6.75$ V.

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

Therefore, $V_D=6.75-2.25=4.5\,\text{V}$ $V_C=4.5-2.25=2.25\,\text{V}$

7. Potential difference across a diagonal

The two diagonals are between:

• $A$ and $C$
• $B$ and $D$

Their potential differences are: $V_{AC}=V_A-V_C=6.75-2.25=4.5\,\text{V}$ $V_{BD}=V_D-V_B=4.5-0=4.5\,\text{V}$

So the potential drop across each diagonal is $4.5\,\text{V}=45\times 10^{-1}\,\text{V}$

8. Final integer answer

The blank is $\boxed{45}$