It would definitely be bullet #2 that hit the glass first... followed by bullet #3 and last was bullet #1. The reasoning being that #2 doesn't have any cracks that are "stopped" by existing cracks, #3 only has a crack stopped by cracks from #2, and #1 has cracks that are stopped by both #2 and #3.

That was a pretty easy one compared to the maths involved with the probabilities of dying in games of Russian roulette