Explanation

(c) Excess of pressure inside a soap bubble
$p = \frac{4T}{R}$
$or\ \ \ \frac{p_{1}}{p_{2}} = \frac{R_{2}}{R_{1}}$
$Given,\ \ \ \ p_{1} = 3p_{2}$
$\therefore\ \ \frac{3p_{2}}{p_{2}} = \frac{R_{2}}{R_{1}}$
$or\ \ \ \ \frac{R_{1}}{R_{2}} = \frac{1}{3}$
Therefore, ratio of volume of bubbles
$\frac{V_{1}}{V_{2}} = \frac{\frac{4}{3}\pi R_{1}^{3}}{\frac{4}{3}\pi R_{2}^{3}}$
$= \frac{R_{1}^{3}}{R_{2}^{3}} = \left( \frac{1}{3} \right)^{3} = \frac{1}{27}$
$\therefore\ \ \ \ V_{1}:V_{2} = 1:27$