Explanation

(d) Excess pressure inside a liquid drop
$\mathrm{\Delta}p = \frac{2T}{R}$
Where T is surface tension and R is radius of liquid drop.
$\therefore\ \ \frac{\mathrm{\Delta}p_{1}}{\mathrm{\Delta}p_{2}} = \frac{R_{2}}{R_{1}} = \frac{0.75}{0.50}$
$\Longrightarrow \ \ \frac{\mathrm{\Delta}p_{1}}{\mathrm{\Delta}p_{2}} = \frac{3}{2}$