Answer
The ISS is 373 km above the surface of the earth.
Work Step by Step
We can find the period $T$ of one orbit.
$T = \frac{(24~hours/day)(3600~s/hour)}{15.65~rev/day}$
$T = 5521~s$
We can find the orbital radius $R$.
$T^2 = \frac{4\pi^2~R^3}{G~M}$
$R^3 = \frac{G~M~T^2}{4\pi^2}$
$R = (\frac{G~M~T^2}{4\pi^2})^{1/3}$
$R = [\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)(5521~s)^2}{4\pi^2}]^{1/3}$
$R = 6753~km$
The distance above the earth's surface is 6753 km - 6380 km, which is 373 km.
The ISS is 373 km above the surface of the earth.