A projectile is thrown straight upward from the earth's surface with an initial speed v = α v E ,…

Escape velocity is given by: $v_E=\sqrt{\frac{2GM}{R}}$

At Earth's surface:

Kinetic energy of the particle 

$$\begin{gathered} K{E}_1=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}m{\alpha }^2{\left(v_E\right)}^2 \\ =\frac{1}{2}m{\alpha }^2\times \frac{2GM}{R} \\ =\frac{GMm{\alpha }^2}{R} \end{gathered}$$

Potential energy of the particle:

$P{E}_1=\frac{-GMm}{R}$

At certain height the particle stops for a moment:

$K{E}_2=0$

$P{E}_2=\frac{-GMm}{R+h}$

Applying conservation of mechanical energy,

$$\begin{gathered} K{E}_1+P{E}_1=K{E}_2+P{E}_2 \\ \Rightarrow \frac{GMm{\alpha }^2}{R}+\frac{-GMm}{R}=0+\left(\frac{-GMm}{R+h}\right) \\ \Rightarrow \frac{1}{R}\left({\alpha }^2-1\right)=\frac{-1}{R+h} \\ \Rightarrow \frac{1}{6400}\left({\alpha }^2-1\right)=\frac{-1}{7200} \\ \Rightarrow \alpha =\frac{1}{3} \end{gathered}$$