Explanation

(d) Excess pressure inside a spherical drop of water
$p = \frac{2T}{R}$
$Given,\ \ \ \ p_{1} = 4p_{2}$
$\frac{2T}{R_{1}} = 4 \times \frac{2T}{R_{2\ }}$
$or\ \ \ R_{2} = 4R_{1}$
$Now,\ \ \frac{m_{1}}{m_{2}} = \frac{4\pi R_{1}^{3}d_{1}}{4\pi R_{2}^{3}d_{2}}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \frac{m_{1}}{m_{2}} = \frac{R_{1}^{3}}{R_{2}^{3}}$
$\frac{m_{1}}{m_{2\ }} = \frac{1}{64}$