1. Checking Decidability :
By Rice's theorem, It can be shown that It is Undecidable because It is a Non-Trivial Property of RE languages.
Any non-trivial property of the LANGUAGE recognizable by a Turing machine (recursively enumerable language) is undecidable. For a property of recursively enumerable set to be non-trivial, there should exist at least recursively enumerable languages, (hence two Turing machines), the property holding for one (Tyes being its TM) and not holding for the other (Tno being its TM).
Lyes = { a, aaa, aaaaa } Lno = { a, aa, aaa, aaaaa} ..Thus There Exists a Language for which the given Property is satisfied and There Exists a Language for which the given Property is Not satisfied. Thus, The Property is Non-trivial and So, The Language is Undecidable.
But Undecidable means NOT REC. Undecidable could be RE or NOT RE.
2. Checking Recognizability :
Now By Using Rice's Extended( Recognizability ) theorem, We can show that the Language is Unrecognizable (i.e. Not RE)
Any non-monotonic property of the LANGUAGE recognizable by a Turing machine (recursively enumerable language) is unrecognizable. For a property of recursively enumerable set to be non-monotonic, there should exist at least two recursively enumerable languages (hence two Turing machines), the property holding for one (Tyes being its TM) and not holding for the other (Tno being its TM) and the property holding set (language of Tyes) must be a proper subset of the set not having the property (language of Tno).
Lyes = { a, aaa, aaaaa } Lno = { a, aa, aaa, aaaaa}
Lyes $\subset$ Lno thus, Language is Not RE.
Thus Option C is correct.
Useful Link : https://gatecse.in/rices-theorem/