OK. Let's take that the question is poorly worded, and work from there. The particle, regardless of velocity, is vectored. If it were not, say it was running directly away from the source, it would slow down, stop, and then fall toward the field. If it was running straight into the source, it would experience an increasing F gradient, but no deflection angle.
The largest deflection in a particle always occurs at right angles between the particle and the source, regardless of relative masses and velocities. (This completely disregards the effect of the forces on the final vecors of the the masses in question, but c'est la vie)
So, if the largest deflection is F at radius 1 (arbitrary, chosen for unit circle and easy lookup in standard tables for cos and sin) is 19 degrees. The question breaks down to this: At what radius from the source (less than 1) does the inverse square law yield a field strength of 2.2. Then, if the particle was at 19 degrees to point 1, what angle is it to point 2?