This seems to be an example of the Look-and-say sequence, which is in itself interesting; its variant, the Kolkoski sequence, leads to several difficult problems.
Specifically, each term of the sequence after the initial one describes the previous term of the sequence by listing the number of times each symbol appears in the term. Thus, since the first term in the sequence is ABC, the next term should tell us that we have one A, one B, and one C, and is therefore 1A1B1C. This term has one 1, one A, one 1, one B, one 1, and one C, so the third term of the sequence is 111A111B111C. Continuing this way, we have 311A311B311C, 13211A13211B13211C, and 111312211A111312211B111312211C, so the answer appears to be option e).
That being said, as it is always the case with these problems, there is nowhere near enough information to actually answer the question in a way that is mathematically justifiable, and any term whatsoever is the sixth term of a sequence that begins as the one you listed. This, of course, is never considered by whoever asks these questions.