Given: Period of satellite $T=6 \mathrm{hrs}=6\times 60\times 60=21600 \mathrm{s}$

Radius of earth $=6400 \mathrm{km}$

Let $h$ be the distance of the satellite from the surface of the earth.

$h=R_{\mathrm{orbit}}-R_{\mathrm{earth}}$, where $R_{\mathrm{orbit}}$ is the radius of orbit of the satellite.

Using Kepler's third law of planetary motion $\frac{T^2}{R_{\mathrm{orbit}}^3}=\frac{4{\pi }^2}{GM}$

$R_{\mathrm{orbit}}=T^2\times \left(\frac{GM}{4{\pi }^2}\right)={\left[{\left(21600\right)}^2\times 8\times {10}^{12}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=3.732\times {10}^{20}\times {10}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Thus, $R_{\mathrm{orbit}}=1.512\times {10}^7 \mathrm{m}$

Hence the distance of the satellite from the surface of the earth is 

$h=1.512\times {10}^7-6.4\times 10=8720 \mathrm{km}$