The question is, if the space ship traveled past earth at near the speed of light, how much acceleration would gravity act upon the space ship?
I'm tempted to ask, "How much color would a voltage of 2.34 V impart on the ship?"
Your terminology is close enough to quibble about, since if one does not quibble a bit, serious misunderstanding is the most likely result.
Newtonian (non-relativistic) gravitational force is \$F = G m_1 m_2 / r^2\$, where \$r\$ is the distance between the point-like masses \$m_1\$ and \$m_2\$, and \$G\$ is
the gravitational constant, \$G = 6.674 \cdot 10^11 \, m^3 \, kg^{-1} \, s^{-2}\$. The velocity of either point-like mass does not affect the force itself, only how the force varies as a function of time. The acceleration of a point like mass towards the center of mass of the pair is given by \$\vec{F} = m \vec{a}\$, so we can express the Newtonian acceleration of a point-like mass \$m\$ due to the gravitational effects of another point-like mass \$M\$ at \$\vec{r}\$ relative to the first point-like mass as
$$\vec{a} = \frac{G \; M}{\lVert \vec{r} \rVert^2} \frac{\vec{r}}{\lVert\vec{r}\rVert} = \frac{G \; M \; \hat{\Delta}}{r^2}$$
in barycentric coordinates (relative to the center of mass of the pair), with \$\vec{r} = r \hat{\Delta}\$, and \$\lVert\hat{\Delta}\rVert = 1\$ being the direction vector from the point-like mass towards the other point-like mass.
Proper relativistic model uses differential calculus to express the relationships. If we approximate those solutions with polynomials, for small masses and small velocities we get
Einstein-Infeld-Hoffman equations:
$$\begin{aligned}
\vec{a} & = \frac{G \; M \; \hat{\Delta}}{r^2} \\
~ & + \frac{1}{c^2} \frac{G \; M \; \hat{\Delta}}{r^2} \left[ \vec{v} \cdot \vec{v} + 2 \vec{V} \cdot \vec{V} - 4 \vec{v} \cdot \vec{V} - \frac{3}{2}(\hat{\Delta}\cdot\vec{V})^2 + \frac{1}{2}\vec{r}\cdot\vec{A} \right] \\
~ & + \frac{1}{c^2} \frac{G \; M}{r^2} \left[ \hat{\Delta} \cdot ( 4 \vec{v} - 3 \vec{V} ) \right] \left[ \vec{v} - \vec{V} \right] \\
~ & + \frac{7}{c^2} \frac{G \; M \; \vec{A}}{r} \\
~ & + O(c^{-4}) \\
\end{aligned}$$
The final term just explicitly tells what we omitted; the polynomial itself is infinite, but successive terms contribute less and less to the result. \$\vec{v}\$ is the velocity of the point-like mass itself, and \$\vec{V}\$ and \$\vec{A}\$ are the velocity and acceleration of the other point-like mass, all in barycentric coordinates (with respect to the center of mass of the system).
Closer to the speed of light not only do we get Lorentz contraction, but relativistic effects of mass also. The effective mass of a point-like particle with velocity \$v\$ is
$$m_{\text{rel}} = \frac{m_{\text{rest}}}{\sqrt{1 - \frac{v^2}{c^2}}}$$
Relativistic time dilation is described by a similar relation,
$$\Delta t_{\text{rel}} = \frac{\Delta t_{\text{rest}}}{\sqrt{1 - \frac{v^2}{c^2}}}$$
and if we have time dilation ratio of about ten, we know that \$v/c \ge \sqrt{99/100} \approx 0.994987\$. Even with respect to the barycenter of the two-point system, their velocities are around half \$c\$ (because one is massive and almost stationary, and the other is light but moving at near lightspeed), causing a relativistic mass \$m_{\text{rel}} \approx \sqrt{4/3} m_{\text{rest}}\$, or \$m_{\text{rel}} \approx 1.1547 m_{\text{rest}}\$.
So, the fact is, the gravitational interaction between the zipping spaceship and the planet is actually
increased due to the velocity, as it increases the effective mass of the spaceship. The higher velocity only means their relative distance \$r\$ is small (and interaction and therefore the acceleration large) for a short period of time.
Thus, the answer is neither of the suggested ones. The force, and the accelerations in the barycentric coordinates, are somewhat larger in magnitude than they would be at very slow, non-relativistic velocities, because of relativistic effects of the effective masses of the spaceship and the planet in the barycentric coordinate system (coordinates with respect to the center of mass of the pair). The time dilation factor itself is not involved here at all.
See how important that quibbling about terminology was?
The repeating qualifier, "in the barycentric coordinate system", isn't very important in this system, because the center of the mass is where one expects it to be in this case. However, if we consider two spaceships traveling at near lightspeed passing by each other, and they have different masses, the picture becomes much more complicated; yet, using the above barycentric coordinates the definitions still hold.
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