We know that at the surface of the earth, value of g = \\dfrac{G M}{R^2} At height h above the Earth's surface, the value of acceleration due to gravity g' = \\dfrac{GM}{(R+h)^2} So it is given that g' = \\dfrac{g}{4} When the bullet reaches maximum height, acceleration due to gravity is \\dfrac{1}{4} th of that at planet's surface. That implies \\dfrac{GM}{4R^2} = \\dfrac{GM}{(R+h)^2} \\ \\ \\rightarrow h = R By conservation of mechanical energy, \\dfrac{-GMm}{R} + \\dfrac{1}{2}mv^2 = \\dfrac{-GMm}{h+R } + 0 since velocity is zero at max height. \\Rightarrow \\dfrac{1}{2}mv^2 = \\dfrac{GMm}{2R} v = \\sqrt{\\dfrac{GM}{R}} = \\sqrt{\\dfrac{2GM}{2R}} = \\dfrac{1}{\\sqrt{2}} \\sqrt{\\dfrac{2GM}{R}}= \\dfrac{1}{\\sqrt{2}}v_{esc} \\Rightarrow v_{esc} = \\sqrt{2}v \\Rightarrow N = 2