= Solution
{c}
For a <timelike circular Schwarzschild geodesic>, the coordinate <orbital period> and onboard <proper time> per orbit are
$$
T_\infty=\frac{2\pi}{\Omega}=2\pi\sqrt{\frac{R^3}{GM}},\qquad
T_{\rm crew}=\int_0^{T_\infty}\frac{dt}{u^0}
=T_\infty\sqrt{1-\frac{3GM}{Rc^2}}.
$$
Using $G=6.67430\times10^{-11}\,{\rm m^3\,kg^{-1}\,s^{-2}}$ and $c=299792458\,{\rm m\,s^{-1}}$, the <Schwarzschild radius> and orbital <areal radius> are $R_s=2GM/c^2\simeq1.21789\times10^{10}\,\mathrm m$ and $R=200R_s\simeq2.43578\times10^{12}\,\mathrm m$. Therefore
$$
\boxed{\begin{aligned}
T_\infty&\simeq1.02100\times10^6\,\mathrm s=11.8172\,\mathrm{days},\\
T_{\rm crew}&=T_\infty\sqrt{\frac{397}{400}}
\simeq1.01717\times10^6\,\mathrm s=11.7728\,\mathrm{days}.
\end{aligned}}
$$
The crew accumulates about $3836$ fewer seconds, or $63.9$ fewer minutes, per orbit. Both <gravitational time dilation> and kinematic <time dilation> contribute: a static clock at $R$ has $dT_{\rm static}=\sqrt{1-R_s/R}\,dt$, while the orbiting clock additionally has $d\tau=dT_{\rm static}/\gamma$, where $\gamma=\sqrt{398/397}$. A distant observer measures the coordinate period; comparing arrivals emitted at the same orbital phase also removes the repeated light propagation delay.
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