You know that:
- A belongs to NP
- B is NP-Complete
- C is NP-Complete
- If A accepts given word then B accepts this word as well
- If we can solve C - we can solve A.
To prove that A is NP-Complete you need to prove that it belongs to NP (which is given to us) and that any problem from NP can be reduced to it (so that it is NP-Hard). To show the second one you need to show a reduction from an NP-Complete problem to A.
The last (5th) of our information basically means you have a polynomial reduction from A to C which is no surprise, since A is in NP and C is NP-Complete (meaning, among others, that every NP problem can be reduced to it, including A). This doesn't give you much about A (it only confirms that A is no harder than C).
The 4th information gives us knowledge like this: if some word belongs to A it also belongs to B, but there are some possible words that belong to B but not to A. In other words, A is a subset of B. This does not gives us much about A either, because A could belong to P only and still be a subset of B (as NP "contains" P).
In my opinion, if I understood this well, this information is not sufficient to say more about the complexity of A.