Your solution is also correct. The LTL formula X X G F[a SU b] is equivalent to the LTL formula G F b, so that the X-operators are not that meaningful here. You may also add further ones, since X ... X G F phi is equivalent to G F phi, since if phi holds infinitely often, it will do so also after some finite time. The other thing that you used and the example solution used as well is that F[a SU b] is equivalent to F b: If F[a SU b] holds at some time t, then there is a point of time t+delta where b holds, and therefore also F b holds. Conversely, if F b holds, then there is a point of time t where b holds, and at that time also [a SU b] holds, since b implies [a SU b].

Still confused?