\(V_\theta=\sqrt{\frac{2 G M}{R}}\)

\(M \rightarrow \text { mass of earth }\)

\(R \rightarrow \text { Radius of earth }\)

Now, conserving potential energy at the surface of earth and highest point,

\(-\frac{G M m}{R}+\frac{1}{2} m\left(\frac{1}{2} \sqrt{\frac{2 G M}{R}}\right)^2=-\frac{G M m}{r}\)

\(-\frac{G M m}{R}+\frac{G M m}{4 R}=-\frac{G M m}{r}\)

\(-\frac{3 G M m}{4 R}=-\frac{G M m}{r}\)

\(\Rightarrow r=\frac{4 R}{3}\)

\(\Rightarrow R+h=\frac{4 R}{3}\)

\(\Rightarrow h=\left(\frac{R}{3}\right)\)