Answer
$h = 3.19\times 10^6~m$
Work Step by Step
The gravitational potential energy at the Earth's surface is $U_s = -\frac{GMm}{R}$
The gravitational potential energy at a height $h$ above the Earth's surface is $U_h = -\frac{GMm}{R+h}$
The kinetic energy of a satellite in orbit at a height of $h$ above the Earth's surface is $K_h = \frac{GMm}{2(R+h)}$
We can find $h$:
$U_s+K_h = U_h$
$-\frac{GMm}{R}+\frac{GMm}{2(R+h)} = -\frac{GMm}{R+h}$
$\frac{GMm}{2(R+h)} +\frac{GMm}{R+h} = \frac{GMm}{R}$
$\frac{1}{2(R+h)} +\frac{2}{2(R+h)} = \frac{1}{R}$
$\frac{3}{2(R+h)} = \frac{1}{R}$
$2(R+h) = 3R$
$2h = R$
$h = \frac{R}{2}$
$h = \frac{6.37\times 10^6~m}{2}$
$h = 3.19\times 10^6~m$
