The approximate height from the surface of earth at which the weight of the body becomes 1 3 of its weight…

The variation of the acceleration due to gravity with height $h$ is given by,

$g^{\text{'}}=\frac{g}{{\left(1+{\displaystyle \frac{h}{R}}\right)}^2}$.

Since the weight at given height is $\frac{1}{3}$ of the weight at the surface

$$\begin{gathered} \Rightarrow g^{\text{'}}=\frac{g}{3} \\ \Rightarrow \frac{g}{{\left(1+{\displaystyle \frac{h}{R}}\right)}^2}=\frac{g}{3} \\ \Rightarrow \left(1+\frac{h}{R}\right)=\sqrt{3} \\ \Rightarrow h=R\left(\sqrt{3}-1\right)=6400\times \left(1.732-1\right)=4685 \mathrm{km} \end{gathered}$$