I think that the authors of this question made a silly mistake, in that they did not include the formal charge. Let's assume for now that it is indeed a carbanion, and not a carbene. To have p orbitals available for resonance at every carbon, the carbanion must be sp2 hybridized. At the same time, doing this does not satisfy Hückel's rule. If we leave the carbon sp3 hyrbridized, then it is not a fully conjugated system. So, assuming it is indeed a carbanion, it is certainly not aromatic.
Taking the case of the carbene, well I don't know enough about carbenes to answer it, honestly. But from the context of the question, I highly doubt the authors would intend for you to consider a carbene.
So, my answer: C