The pressures inside two soap bubbles A and B are $1.02$ atmosphere and $1.04$ atmosphere respectively. The ratio of volume of bubble A to that of bubble B is (outside pressure $= 1$ atmosphere)

The excess pressure inside a soap bubble is given by $\Delta P = \dfrac{4T}{R}$, where $T$ is the surface tension and $R$ is the radius of the bubble.

For bubble A, the excess pressure is $\Delta P_A = 1.02 - 1 = 0.02$ atm.

For bubble B, the excess pressure is $\Delta P_B = 1.04 - 1 = 0.04$ atm.

Taking the ratio of the excess pressures, we get $\dfrac{\Delta P_A}{\Delta P_B} = \dfrac{0.02}{0.04} = \dfrac{1}{2}$.

Since $R \propto \dfrac{1}{\Delta P}$, the ratio of their radii is $\dfrac{R_A}{R_B} = \dfrac{\Delta P_B}{\Delta P_A} = \dfrac{2}{1}$.

The volume of a spherical bubble is $V = \dfrac{4}{3}\pi R^3$, which means $V \propto R^3$.

The ratio of their volumes is $\dfrac{V_A}{V_B} = \left(\dfrac{R_A}{R_B}\right)^3 = (2)^3 = 8$.

Thus, the ratio of the volume of bubble A to that of bubble B is $8:1$.

Answer: $8:1$