Key Idea Horizontal ranges are same for complementary angles of projection ie, for $\theta$ and $\left(90^{\circ}-\theta\right)$
We know that if two stones have same horizontal range, then this implies that both are projected at $\theta$ and $90^{\circ}-\theta$
Here, $\theta=\frac{\pi}{3}=60^{\circ}$
$\therefore 90^{\circ}-\theta=90^{\circ}-60^{\circ}=30^{\circ}$
For first stone, Max. height $=102=\frac{u^{2} \sin ^{2} 60^{\circ}}{2 g}$
For second stone, Max. height, $h=\frac{u^{2} \sin ^{2} 30^{\circ}}{2 g}$
$\therefore \frac{h}{102}=\frac{\sin ^{2} 30^{\circ}}{\sin ^{2} 60^{\circ}}=\frac{(1 / 2)^{2}}{(\sqrt{3} / 2)^{2}}$
or $h=102 \times \frac{1 / 4}{3 / 4}=34 \,m$