Two isolated solid metallic spheres A and B have radii in the ratio...

The electric potential at the surface of an isolated solid metallic sphere is given by $V = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{R}$.

For spheres $A$ and $B$, the ratio of their potentials is:

$\dfrac{V_A}{V_B} = \dfrac{Q_A}{Q_B} \times \dfrac{R_B}{R_A}$

Given $\dfrac{R_A}{R_B} = \dfrac{1}{2}$ and $\dfrac{Q_A}{Q_B} = \dfrac{3}{4}$, substituting these values:

$\dfrac{V_A}{V_B} = \dfrac{3}{4} \times \dfrac{2}{1} = \dfrac{3}{2}$

The ratio of the electric potentials is $3 : 2$.

Answer: $3 : 2$