Answer
All six paths from point a to point b result in the same net work done on the rocket by the gravitational force from the moon.
Work Step by Step
Let $M$ be the mass of the moon.
Let $m$ be the mass of the rocket.
We can write an expression for the gravitational potential energy at point $a$:
$U_a = -\frac{GMm}{r_a}$
We can write an expression for the gravitational potential energy at point $b$:
$U_b = -\frac{GMm}{r_b}$
We can write an expression for the change in the gravitational potential energy from point a to point b:
$\Delta U = U_b-U_a$
$\Delta U = -\frac{GMm}{r_b}-(-\frac{GMm}{r_a})$
$\Delta U = \frac{GMm}{r_a}-\frac{GMm}{r_b}$
$\Delta U = GMm~(\frac{1}{r_a}-\frac{1}{r_b})$
Note that the change in gravitational potential energy only depends on the initial position and the final position. It does not depend on the path.
Therefore, all six paths from point a to point b result in the same change in gravitational potential energy.
We can write an expression for the net work done on the rocket by the gravitational force from the moon:
$W_g = -\Delta U$
Since all six paths from point a to point b result in the same change in gravitational potential energy, then all six paths from point a to point b result in the same net work done on the rocket by the gravitational force from the moon.
