Explanation

(c) $H = \frac{u^{2}\sin^{2}\theta}{2g}\ \Longrightarrow \frac{H_{1}}{H_{2}} = \frac{u^{2}\sin^{2}\theta_{1}}{u^{2}\sin^{2}\theta_{2}}$
$\frac{3}{1} = \frac{\sin^{2}\theta_{1}}{\sin^{2}\theta_{2}}\ \Longrightarrow \frac{\sin\theta_{1}}{\sin\theta_{2}} = \frac{\sqrt{3}}{1}$
Logically, we can conclude that
$\theta_{1} = 60{^\circ},\ \ \theta_{2} = 30{^\circ}$
$Again\ R = \frac{u^{2}sin2\theta}{g}$
$\therefore\frac{R_{1}}{R_{2}} = \frac{4\ u^{2}sin2\theta_{1}}{u^{2}sin2\theta_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{4sin2(60{^\circ})}{sin2(30{^\circ})} = \frac{4sin120{^\circ}}{sin60{^\circ}}$
$or\ \ \frac{R_{1}}{R_{2}} = \frac{4 \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = 4\ $