= 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 $86164\,\mathrm s$. Thus
$$
\boxed{r_{\rm geo}=6550\left(\frac{86164}{87\times60}\right)^{2/3}\,\mathrm{km}
\simeq4.25\times10^4\,\mathrm{km}.}
$$
Its altitude is about $3.61\times10^4\,\mathrm{km}$ 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.
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