 Solution

For two circular Kepler orbits about the same body, Kepler's third law gives $T^2/r^3=\text{constant}$. The reference radius is measured from Earth's centre:A geostationary orbit is prograde and equatorial, with one sidereal day as its orbital period. Using $T_g\simeq 1436\min$ and the supplied $T_1=109\min$,Using $24\mathrm{h}$ instead changes this estimate by less than one per cent. This is the orbital radius, not the altitude; the corresponding altitude is about $3.00\times 10^4\mathrm{km}$.

There is a numerical inconsistency in the source data. The quoted reference orbit implies $G{M}_E=4{\pi }^2{r}_1^3/T_1^2\simeq 2.56\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, appreciably below the measured terrestrial value, about $3.99\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$. Using the latter instead gives the familiar geostationary orbit radius $4.22\times 10^4\mathrm{km}$. The boxed estimate consistently uses the period actually printed in the question.