Answer: B
The ketone takes one CHX3MgBr to give a tertiary alkoxide carbon bearing two methyls (C(CHX3)X2OH end). The ester takes two equivalents (via a ketone) to give a second C(CHX3)X2OH end. The product is (CHX3)X2C(OH)CHX2CHX2C(OH)(CHX3)X2, using 3 mol of Grignard reagent.