If g is the acceleration due to gravity at the earth's surface and r is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is

Let the mass of the body be $m$. When the body will escape earth's gravity into space it will have kinetic and potential energies as zero.

$(KE+PE{)}_{surface}=(KE+PE{)}_{space}$

$\Rightarrow \frac{1}{2}m{v}_e^2-mgr=0$ ( $r$ is the radius of earth)

$\Rightarrow \frac{1}{2}m{v}_e^2=mgr$

$\Rightarrow v_e^2=2gr$.$\Rightarrow v_e=\sqrt{2gr}$ is our required answer,