We use conservation of energy to find the speed of the stone. All the energy stored in the elastic bands is transferred to the stone as kinetic energy.

Energy stored in the elastic bands is $$E =2 \frac{k\left(\sqrt{x^2 + \left(\sqrt{L^2 - H^2}\right)^2} - \sqrt{L^2 - H^2}\right)^2}{2}$$ because we have two of them and the amount by which one is extended is $$\sqrt{x^2 + \left(\sqrt{L^2 - H^2}\right)^2} - \sqrt{L^2 - H^2}\;.$$

Suppose that the initial speed of the stone is $v$ then it has a kinetic energy $E = \frac{mv^2}{2}$. The stone is fired at an angle of $45^\circ$. The initial speed of a projectile and the distance it travels is related by $d = \frac{v^2}{g}$. It is given that $d = 50\mathrm{m}$. Thus, the kinetic energy is $E = \frac{mgd}{2}$.

By the conservation of energy $${k\left(\sqrt{x^2 + L^2 - H^2} - \sqrt{L^2 - H^2}\right)^2} = \frac{mgd}{2}\;.$$ Now, the numbers $L = 0.12\mathrm{m}$ , $H = 0.10\mathrm{m}$ , $m = 0.05\mathrm{kg}$ , $k = 200\mathrm{N/m}$ , $d = 50\mathrm{m}$, $g = 9.81\mathrm{m/s^2}$ could be plugged in to find $x$ or we can simplify to get $$x = \sqrt{\frac{mgd}{2k} + \sqrt{(L^2 - H^2) \frac{2mgd}{k}}}\;.$$

Thus, we have that $x = 31\mathrm{cm}$.