The height ‘h’ at which the weight of a body will be the same as that of at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected):

g at height  $h=\frac{G{M}_e}{{\left(R_e+h\right)}^2}$
g at depth  $d=\frac{G{M}_e}{R_e^2}\left(1-\frac{d}{R_e}\right)$
For Weight to be equal  $\frac{G{M}_e}{R_e^2{\left(1+\frac{h}{R_e}\right)}^2}=\frac{G{M}_e}{R_e^2}\left(1-\frac{d}{{\mathrm{R}}_e}\right)$
Given d = h  
$\begin{array}{l}\Rightarrow 1={\left(1+\frac{h}{R_e}\right)}^2\left(1-\frac{h}{R_e}\right)\text{}\\ \Rightarrow h^2+hR-R^2=0 \text{}\\ comparing with a{x}^2+bx+c=0\\ a=1,b=R and c=-R^2\text{}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \text{}h=\frac{-R\pm \sqrt{R^2-4\left(1\right)\left(-R^2\right)}}{2\left(1\right)}\\ h=\frac{-R\pm \sqrt{R^2+4R^2}}{2}\text{}\\ \Rightarrow h=\frac{\sqrt{5}R-R}{2}\end{array}$