= Solution
The transfer is a <Hohmann transfer orbit>. Write $r_0$ for the original <circular Kepler orbit> radius and $a_t$ for the transfer <ellipse>'s <semi-major axis>. The diagram measures both radii from <Earth>'s centre, giving $r_1=1.2r_0$, and hence
$$
a_t=\frac{r_0+r_1}{2}=1.1r_0.
$$
The <Hohmann transfer time> is half the transfer <orbital period>, taking the satellite to the opposite end of the <elliptic orbit>. By <Kepler's third law>, relative to the original <geostationary orbit>'s <orbital period> $T_0$,
$$
\frac{T_t}{T_0}=\left(\frac{a_t}{r_0}\right)^{3/2},\qquad
\boxed{t=\frac{T_0}{2}(1.1)^{3/2}\simeq13.8\ \mathrm h.}
$$
Using one <sidereal day>, $T_0=86164\ \mathrm s$, gives $t=4.9703\times10^4\ \mathrm s=13.806\ \mathrm h$; the usual $24\ \mathrm h$ approximation gives $13.844\ \mathrm h$. The arrival impulse changes the <velocity> into that of the new <circular Kepler orbit> without adding coast time in the impulsive model.
There is a real discrepancy between the wording and the diagram: an increase in altitude above <Earth>'s surface is different from the radius increase drawn. If altitude is interpreted literally, let $R_E$ be <Earth>'s radius. Then
$$
r_1=R_E+1.2(r_0-R_E),\qquad
\boxed{t=\frac{T_0}{2}\left(1.1-0.1\frac{R_E}{r_0}\right)^{3/2}\simeq13.5\ \mathrm h,}
$$
using $R_E\simeq6371\ \mathrm{km}$ and $r_0\simeq42164\ \mathrm{km}$. Thus $13.8\ \mathrm h$ answers the explicitly labelled geometry; $13.5\ \mathrm h$ answers the literal altitude wording.
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