Question

at what depth from the surface of earththe acceleration due to gravity will become 1/4th the value of g on surface

Answer 1

gravity at a depth d is given by

$g(d) = GM_e \frac{R_e-d}{R_e^3}$

given that gravity at depth d is 1/4th the value of g on surface. So

$\frac{g(d)}{g(0)}= \frac{1}{4}\\ \\ \Rightarrow \frac{GM_e \frac{R_e-d}{R_e^3} }{\frac{GM_e}{R_e^2} } = \frac{1}{4} \\ \\ \Rightarrow \frac{R_e-d}{R_e^3} \times R_e^2 = \frac{1}{4} \\ \\ \Rightarrow \frac{R_e-d}{R_e}= \frac{1}{4} \\ \\ \Rightarrow 4R_e-4d = R_e\\ \\ \Rightarrow d = \frac{3}{4}R_e$

So at a depth of 3R/4, gravity will be 1/4th of the value of g on surface.