Answer
$V_0=\sqrt { 2G M_e(\dfrac{1}{R_e}-\dfrac{1}{r+R_e})}$$~m/s$
Work Step by Step
The work needed to move the rocket a distance $r$ from the surface of the earth can be calculated as:
$ Work \ done =G M_e \ m (\dfrac{1}{R_e}-\dfrac{1}{r+R_e})$
The rocket will attain its maximum height when its kinetic energy reduces to $0$.
That is,
$\dfrac{1}{2} m V_0^2 =G M_e \ m (\dfrac{1}{R_e}-\dfrac{1}{r+R_e})$
Now, the initial velocity will be
$V_0=\sqrt {\dfrac{2G M_e \ m }{m} (\dfrac{1}{R_e}-\dfrac{1}{r+R_e})}$
or, $V_0=\sqrt { 2G M_e(\dfrac{1}{R_e}-\dfrac{1}{r+R_e})}$
