Nobody in this forum would like to spoil your pleasure to solve such a nice exercice by yourself.
Therefore, nobody will give you the answer.
You have to find out where a beam from the source to the detector should cross the interface.
The Fermat principle tells you that this should be the place such that the time travel will be the shortest.
To save you a little bit of time, ask yourself first if this crossing point should be more on the side of the source or on the side of the detector.
Observe that if this point is more on the side of the source, the "optical" (seismic) path will be shorter on the side of the source where the wave is also the slower (2000 m/s) and it will be longer on the other side.
Intuitively, I would expect this point to be on the side of the source.
You could calculate the flight time for different positions and find out where it is minimum.
I checked that the crossing point is very close to a grid point on your drawing.
You will check that the total filght time is 710 ms, then.