I think it is very unfortunate if you need to do something that is wrong in order to pass a course. I would replace the gravitational force with an electromagnetic force with and state this in the beginning, but that is just me. It also unfortunately depends on how you think your professor would react.
Now, to the actual approach to this in GR, the problem is not well formulated because it is not clear what would be meant by ##\gamma##. There are a few possibilities that include (but I probably one can think of others too):
- The ##t##-component of the 4-velocity of the particle, i.e., ##\dot t = dt/d\tau## where ##\tau## is the proper time and ##t## the time coordinate.
- The ##\gamma## factor of the particle as measured by a static observer measuring its speed as it passes by (call this ##\gamma_o##).
These two will give different final results. I would tend to favour number 2 here as it is more physical.
In the Schwarzschild solution (i.e., the spherically symmetric vacuum solution outside a spherically symmetric mass distribution), there is a constant of motion that is ##(1-2GM/(c^2r)) \dot t = \gamma_\infty##, where ##\gamma_\infty## would be the ##\gamma## factor at infinity. This would also mean that, for ##GM/(c^2 r) \ll 1##, you would have
$$
\dot t \simeq \gamma_\infty \left( 1 + \frac{2GM}{c^2 r}\right),
$$
which differs from the result of your problem by a factor ##2\gamma_\infty##. However, as I said, my preferred interpretation would be 2, which would also require taking into account the gravitational time dilation of the static observer, leading to
$$
\gamma_o = \dot t \left( 1 - \frac{2GM}{c^2 r}\right)^{-1/2} = \gamma_\infty \left( 1 - \frac{2GM}{c^2 r}\right)^{-3/2} \simeq \gamma_\infty \left( 1 + \frac{3GM}{c^2 r}\right),
$$
which differs to the result of your problem by a factor ##3\gamma_\infty##. (Assuming I did the math correctly here, it was kind of hastily between other tasks ...)