= Solution
Use the supplied planet <mass> $5.9\times10^{24}\,\mathrm{kg}$, $G=6.67430\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$, and adopt mean Earth radius $R_E=6371\,\mathrm{km}$. For a <geostationary orbit>, use one <sidereal day>, $T\simeq86164\,\mathrm s$. The adopted radius and rotation period follow NASA's fact sheet.<Earth Fact Sheet by National Aeronautics and Space Administration>[Bulk parameters: volumetric mean radius6371.000km; orbital parameters: sidereal rotation period23.9345hours]{ref}
$$
r_0=R_E+300\,\mathrm{km}=6.671\times10^6\,\mathrm m,\qquad r_1=\left(\frac{\mu T^2}{4\pi^2}\right)^{1/3}\simeq4.1994\times10^7\,\mathrm m.
$$
The slightly smaller radius than the usual terrestrial value follows from using the question's smaller Earth <mass>. For coplanar prograde orbits, the preceding formulas give
$$
\Delta v_1\simeq2.410\,\mathrm{km\,s^{-1}},\qquad\Delta v_2\simeq1.459\,\mathrm{km\,s^{-1}},\qquad\boxed{\Delta v_H\simeq3.869\,\mathrm{km\,s^{-1}}.}
$$
The escape transfer has $a\to\infty$, so its first tangential burn reaches the local <escape speed> $\sqrt{2\mu/r_0}$. No final circularization at infinity is needed:
$$
\boxed{\Delta v_{\mathrm{escape}}=(\sqrt2-1)\sqrt{\frac\mu{r_0}}\simeq3.182\,\mathrm{km\,s^{-1}}.}
$$
Thus the geostationary transfer requires about $0.687\,\mathrm{km\,s^{-1}}$ more total impulse, or about 22 percent more, despite leaving the spacecraft gravitationally bound. Reaching the outer radius and circularizing there are separate requirements. Changing the starting orbit's inclination would add a plane-change cost absent from this calculation.
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