ok, so i don't recall any "method" to do this problem, so if someone has a method, please post it. i attempted to use common sense (which i don't do often) so let's see what happens.

if the bug starts at 1, it will jump counter clockwise 2 spaces and end up at 4, since 4 is even, it jumps once to 3, then since 3 is odd, it jumps 2 spaces to one. and then the same process repeats. 1 4 3 1 4 3 1 4 3 1 ....

the bug takes 3 jumps to complete one cycle, that is, after 3 jumps, it is back on point 1 and start over. now, 2007 is divisible by 3. 2007/3 = 669. so the bug will complete exactly 669 cycles after 2007 jumps, and therefore it will be back to point 1 on it's 2007th jump