Solution
Let $V(L)$ be the potential at the center of a cube of side $L$ and uniform charge density $\rho$. By dimensional analysis ($V \approx kQ/L$ and $Q \approx \rho L^3$), the potential scales as:
$$V \propto L^2$$Consider a large cube of side $2L$. Its center potential is $V_{center}(2L)$.
$$V_{center}(2L) = 4 \times V_{center}(L) = 4 V_0$$Now, decompose the large cube (side $2L$) into 8 smaller cubes of side $L$. The center of the large cube corresponds to the common corner of these 8 small cubes.
By superposition, the potential at the center of the large cube is the sum of the potentials at that point due to each of the 8 small cubes. Since the point is a corner for all 8 small cubes:
$$V_{center}(2L) = 8 \times V_{corner}(L)$$Equating the two expressions:
$$4 V_0 = 8 V_{corner}$$ $$V_{corner} = \frac{4 V_0}{8} = \frac{V_0}{2}$$Answer: (b) V₀/2