If any of the sons has cataracts and the parents are healthy then the parents must be heterozygotes. Now if the twins are dizygotic the probability that both sons will have cataracts is 1/16, the probability that only one son will have cataracts is 3/8, and the probability that both sons will be healthy is 9/16. If however the twins are monozygotic then the probability that both sons will have cataracts is 1/4, the probability that only one son will have cataracts is 0, and the probability that both sons will be healthy is 3/4.
Then P(monozygotic | both sons have cataracts) = P(both sons have cataracts | monozygotic) / P(both sons have cataracts) and by Bayes' theorem:
P(monozygotic | both sons have cataracts) = 1/4 / (1/4 * P(monozygotic) + 1/16 * P(dizygotic)) and by pattern matching:
P(dizygotic | both sons have cataracts) = 1/16 / (1/4 * P(monozygotic) + 1/16 * P(dizygotic))
so that it's 4 times more likely that they are monozygotic. I was able to do these deductions and calculations almost instantly despite not being an expert in genetics, so you should be able to figure it out if you have decent mathematical aptitude.
Edit: Actually since it's located on the X chromosome and the mother is healthy the sons will always be healthy unless there are mutations... so the question makes no sense.