d
As the satellite always remains stationary \(w.r.t\) earth surface, thus its time period revolution is equal to time period of rotation of earth i.e \(24\, hrs\)

Time period of satellite \(T=2 \pi \sqrt{\frac{r^{3}}{g R^{2}}}\) where \(R=6400 \mathrm{km}=6.4 \times 10^{6} \mathrm{m}\)

\(\therefore 24 \times 3600=2 \pi \sqrt{\frac{r^{3}}{9.8\left(6.4 \times 10^{6}\right)^{2}}}\)

OR \(\frac{r^{3}}{401.408 \times 10^{12}}=1.89 \times 10^{8} \quad \Longrightarrow r^{3}=76 \times 10^{21}\)

\(\Rightarrow r=42400 \mathrm{km}\)

Thus height of satellite above earth surface \(\quad h=42400-6400=36000 \mathrm{km}\)