= Solution
{c}
For a prograde equatorial <synchronous circular orbit>, equate <gravitational acceleration> and centripetal <acceleration>. With planetary radius $R_M$, $GM=g_sR_M^2$, and rotation period $P$,
$$
\frac{GM}{r_o^2}=\left(\frac{2\pi}P\right)^2r_o,\qquad
r_o=\left[\frac{g_sR_M^2P^2}{4\pi^2}\right]^{1/3}.
$$
The tabulated $g_s$ is a <gravitational acceleration>; its printed unit must be read as $\mathrm{m\,s^{-2}}$, rather than $\mathrm{m\,s^{-1}}$. Using $P=88642\,\mathrm s$ gives
$$
r_o\simeq20398.7\,\mathrm{km},\qquad
\boxed{h=r_o-R_M\simeq17008.7\,\mathrm{km}.}
$$
The <orbital altitude> is measured from the surface, rather than the centre.
At the visible horizon, the line of sight is tangent to the spherical surface. The limiting central angle $\theta_h$ satisfies $\cos\theta_h=R_M/r_o$. The visible <spherical cap> has area $2\pi R_M^2(1-\cos\theta_h)$, giving
$$
\boxed{\frac{A_{\rm vis}}{4\pi R_M^2}=\frac12\left(1-\frac{R_M}{r_o}\right)\simeq0.4169.}
$$
About $41.7\%$ of the whole surface is geometrically visible, with $\theta_h\simeq80.4^\circ$. This is a surface-area fraction, not an apparent-disc area ratio. Useful imaging near the horizon is further limited by foreshortening and the long atmospheric path.
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