= 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:
$$
r_1=(6400+120)\ \mathrm{km}=6520\ \mathrm{km}.
$$
A <geostationary orbit> is prograde and equatorial, with one <sidereal day> as its <orbital period>. Using $T_g\simeq1436\ \mathrm{min}$ and the supplied $T_1=109\ \mathrm{min}$,
$$
\boxed{r_g=r_1\left(\frac{T_g}{T_1}\right)^{2/3}
\simeq3.64\times10^4\ \mathrm{km}
=3.64\times10^7\ \mathrm m.}
$$
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 $3.00\times10^4\ \mathrm{km}$.
There is a numerical inconsistency in the source data. The quoted reference orbit implies $GM_E=4\pi^2r_1^3/T_1^2\simeq2.56\times10^{14}\ \mathrm{m^3\,s^{-2}}$, appreciably below the measured terrestrial value, about $3.99\times10^{14}\ \mathrm{m^3\,s^{-2}}$. Using the latter instead gives the familiar <geostationary orbit> radius $4.22\times10^4\ \mathrm{km}$. The boxed estimate consistently uses the period actually printed in the question.
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