1. Principle used: Potentiometer balance condition

   When the galvanometer shows zero deflection, the potential difference between the two connected points is equal.

   So, if point $P$ is the balancing point on the potentiometer wire $AB$, then $V_{AP} = E$ where $E$ is the emf/potential difference being balanced by the external circuit shown in the figure.

2. Potential gradient along wire $AB$

   The potentiometer wire $AB$ has total length $L = 72\text{ cm}$ and is connected across a source of voltage $V$, so the potential drop along the full wire is $V$.

   Hence the potential gradient is $k = \frac{V}{72}\;\text{V/cm}$

   Therefore, potential drop from $A$ to a point $P$ at distance $x$ cm is $V_{AP} = kx = \frac{V}{72}x$

3. Balance condition from the given circuit

   From the figure, the effective potential difference to be balanced is $\frac{2V}{3}$ Therefore, at null deflection, $\frac{V}{72}x = \frac{2V}{3}$

4. Solve for $x$

   Cancelling $V$ from both sides, $\frac{x}{72} = \frac{2}{3}$ $x = 72\times \frac{2}{3} = 48$

5. Nearest integer

   $x = 48\text{ cm}$

Comparison with stored correct answer

Stored correct answer = $48$

My derived answer is also $48$, so they agree.