= Solution
The original <circular Kepler orbit> has $v_0^2=GM/r_0$. For the transfer <ellipse>, set $r_p=r_0$, $r_a=3r_0$ in part (c):
$$
J_t^2=\frac32GMm^2r_0,\qquad
v_p=\frac{J_t}{mr_0}=\sqrt{\frac32}\,v_0.
$$
The first required speed increase is $\Delta v_1=(\sqrt{3/2}-1)v_0$. Under the question's negligible-mass-change fuel model, fuel use is proportional to the magnitude of the <impulse>. Since the full fuel supply corresponds to $mv_0$,
$$
\boxed{f_1=\frac{m\Delta v_1}{mv_0}=\sqrt{\frac32}-1
=0.2247\simeq22.5\%.}
$$
This is the stated “about a quarter” of the original fuel supply. It is an instance of the <Hohmann transfer impulse fractions>; no variable-mass rocket model is required under the stipulated assumptions.
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