C

To find the distance of a geostationary satellite from the surface of the Earth, we start with the formula for the orbital radius of a satellite in geostationary orbit:
1. Gravitational force equals centripetal force:
$\frac{G M m}{r^2} = \frac{mv^2}{r}$
2. For geostationary satellites, $v = \frac{2\pi r}{T}$ where T is the orbital period (24 hours).
Substituting for $v$ gives us:
$\frac{G M}{r^2} = \frac{4\pi^2 r}{T^2}$
3. Rearranging gives:
$r^3 = \frac{G M T^2}{4\pi^2}$
4. Plugging in the known values:
$G = 6.674 \times 10^{-11} \mathrm{Nm^2/kg^2}$, $M \approx 5.97 \times 10^{24} \mathrm{kg}$, and $T = 86400 \mathrm{s}$ (24 hours) gives us:
Calculate $r, r \approx 4.224 \times 10^7 \mathrm{m}$ (42,241 km).
5. Finally, the distance from the Earth's surface:
Earth's radius $R \approx 6.371 \times 10^6 \mathrm{m}$ thus:
Distance from Earth's surface = $r - R \approx 42,241 km - 6,371 km \approx 35,870 km$.
Therefore, the answer corresponding to this distance is Option C.