Answer
$ v=\sqrt{\frac{GM_E}{R_E+h}}$
Work Step by Step
In the given scenario
$F_{centripetal}=F_{gravity}$
$\implies m(\frac{v^2}{R_E+h})=G\frac{M_E\space m}{(R_E+h)^2}$
This can be rearranged as:
$v^2=\frac{GM_E}{R_E+h}$
$\implies v=\sqrt{\frac{GM_E}{R_E+h}}$