 Solution

Use a circular Kepler orbit to eliminate the terrestrial gravitational parameter: $T^2/r^3$ is constant for both satellites. The reference orbital radius is $r_1=(6400+150)\mathrm{km}=6550\mathrm{km}$, measured from Earth's centre. A geostationary orbit is circular, equatorial and prograde, with the sidereal rotation period, approximately $86164\mathrm{s}$. ThusIts altitude is about $3.61\times 10^4\mathrm{km}$ above the stated surface. The approximate low-orbit input sets the accuracy; replacing it by a modern measured gravitational parameter would answer a different numerical estimate. A $24$-hour approximation changes the radius by less than $0.2\%$. Matching the period alone would not make an inclined or eccentric orbit stationary above one point.