Outside pressure = 1 atm

Pressure inside first bubble = 1.01 atm

Pressure inside second bubble = 1.02 atm

Excess pressure $\Delta P_{1} = 1.01 - 1 = 0.01$ atm

Excess pressure $\Delta P_{2} = 1.02 - 1 = 0.02$ atm

$\Delta P \propto \frac{1}{r} \Rightarrow r \propto \frac{1}{\Delta P} \Rightarrow \frac{r_1}{r_2} = \frac{\Delta P_2}{\Delta P_1} = \frac{0.02}{0.01} = \frac{2}{1}$

Since $V = \frac{4}{3} \pi r^{3} \quad : \therefore \quad \frac{V_{1}}{V_{2}} = \left(\frac{r_{1}}{r_{2}}\right)^{3} = \left(\frac{2}{1}\right)^{3} = \frac{8}{1}$