Consider two charged metallic spheres $$S_1$$ and $$S_2$$ of radii $$R_1$$ and $$R_2$$, respectively. The electric fields $$E_1$$ (on $$S_1$$) and $$E_2$$ (on $$S_2$$) on their surfaces are such that $$\frac{E_1}{E_2} = \frac{R_1}{R_2}$$. Then the ratio $$V_1$$ (on $$S_1$$)/$$V_2$$ (on $$S_2$$) of the electrostatic potentials on each sphere is:

For an isolated charged metallic sphere of radius $$R$$ carrying charge $$Q$$, the electrostatic potential on its surface (with respect to infinity) is given by the well-known formula

$$V \;=\; \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{R}.$$

The electric field just outside the surface of the same sphere is obtained from Coulomb’s law and is

$$E \;=\; \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{R^{2}}.$$

Dividing the second expression by the first, we eliminate the common constant $$\dfrac{1}{4\pi\varepsilon_0}$$ and the charge $$Q$$:

$$\dfrac{E}{V} \;=\; \dfrac{\dfrac{Q}{R^{2}}}{\dfrac{Q}{R}} \;=\; \dfrac{1}{R}.$$

Re-arranging this simple relation, we obtain a very useful connection between the electric field and the potential of a spherical conductor:

$$E \;=\; \dfrac{V}{R}\quad\text{or equivalently}\quad V = E\,R.$$

Now we apply this result separately to the two spheres $$S_1$$ and $$S_2$$. For sphere $$S_1$$ we have $$V_1 = E_1\,R_1$$, and for sphere $$S_2$$ we have $$V_2 = E_2\,R_2$$. Taking the ratio of the two potentials, we write

$$\dfrac{V_1}{V_2} \;=\; \dfrac{E_1\,R_1}{E_2\,R_2} \;=\; \left(\dfrac{E_1}{E_2}\right)\!\left(\dfrac{R_1}{R_2}\right).$$

The problem statement tells us that the electric fields on the two spheres satisfy

$$\dfrac{E_1}{E_2} \;=\; \dfrac{R_1}{R_2}.$$

Substituting this given ratio into the previous expression, we have

$$\dfrac{V_1}{V_2} \;=\; \left(\dfrac{R_1}{R_2}\right)\!\left(\dfrac{R_1}{R_2}\right) \;=\; \left(\dfrac{R_1}{R_2}\right)^{2}.$$

Therefore, the ratio of the electrostatic potentials on the two spheres is the square of the ratio of their radii.

Hence, the correct answer is Option B.