While jumping he converts his initial \\\\(kinetic\\\\) \\\\(energy\\\\) to \\\\(potential\\\\) \\\\(energy\\\\)\\\\(\\dfrac{mv^2}{2}=mgh\\\\)\\\\(v=\\sqrt{2gh}\\\\)Now this \\\\( velocity\\\\) has to be equal to \\\\(escape\\\\) \\\\(velocity\\\\) on the sphere\\\\(Escape\\\\) \\\\(velocity=\\sqrt{\\dfrac{2GM_s}{R}}=\\sqrt{\\dfrac{8 \\pi G R_s^2 \\delta}{3}}\\\\)By comparing both the velocities we get \\\\(Radius(R_s)=\\sqrt{\\dfrac{3gh}{4 \\pi G \\delta}}\\\\)