The given answer is 1,1-dicyclopropylmethanone:
Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five-membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.
Why not make a more stable five-membered ring in the first step, and then again perform E2 to get 2-ethenylcyclopentan-1-one?