Before trying to find the apoapsis of a geostationary orbit, first remember what a geostationary orbit and apoapsis are. Can a path always at the same height have a highest point? No. They are all equal! The speed necessary to complete a revolution every 20 hours is dπ/ 20 per hour, where d is the distance from the object. The orbital speed necessary at any point for a geostationary orbit is that at which centrifugal force equals gravity. Centrifugal force’s acceleration equals 4π2dv/sm outward. And that of gravity equals Gm/d2 inward. We are left with a system of equations that has too many units. We will solve it in Ubiquitous Units.
\vec v=\frac{\vec d\pi}{72000s}\\[8pt] \frac{4\pi\vec v\vec d}{sm}=\frac{Gm_1}{\vec d^2}\\[8pt] G=\frac{6.67408m^3}{10^{11}s^2kg}\\[8pt]