A body of mass m is projected with velocity λ v e in vertically upward direction from the surface of…

Escape velocity is given by, $v_e=\sqrt{\frac{2G{M}_e}{R_e}}$

If the body goes to the height $h$ from the surface of the earth, using mechanical energy conservation,

$\Rightarrow \frac{1}{2}m{\left(\lambda \sqrt{\frac{2G{M}_e}{R_e}}\right)}^2-\frac{G{M}_{e}m}{R_e}=-\frac{G{M}_{e}m}{R_e+h}$

$\Rightarrow {\lambda }^2\frac{G{M}_{e}m}{R_e}-\frac{G{M}_{e}m}{R_e}=-\frac{G{M}_{e}m}{R_e+h}$

$\Rightarrow \left(R_e+h\right)=\frac{R_e}{1-{\lambda }^2}$