Nice one, this is a set-up-two-equations-and-let-algebra-do-the-work question.
1. Let residents = 3k and non-residents = 5k right now, since the ratio is 3:5.
2. Statement 1: add six residents, and the ratio flips to 1:1. So 3k + 6 = 5k, which gives 6 = 2k, k = 3. Non-residents = 5k = 15. That's a full number, so sufficient.
3. Statement 2: three people who were logged as residents actually should have been non-residents. So residents become 3k - 3, non-residents become 5k + 3, and that new ratio is 1:3. Set it up: 3(3k - 3) = 5k + 3. That's 9k - 9 = 5k + 3, so 4k = 12, k = 3. Same answer, non-residents = 15. Also sufficient on its own.
Answer: D, each statement alone gets you there.
What I like about this one is that both statements are really just "correct the count and set the new ratio equal," and once you're comfortable translating word problems into a single variable k for the ratio unit, both statements resolve fast. Early on I used to rewrite R and N as two separate variables and solve a messier system. One variable for the ratio unit does the whole job here, and it's worth practicing that shortcut deliberately instead of hoping it shows up on its own under time pressure.