Answer: B
Br⋅ adds to C-3, the carbon bearing one H, to give the tertiary radical (CHX3)X2C⋅−CHBrCHX3. H abstraction from HBr takes place from either face, giving (CHX3)X2CH−CHBrCHX3, which has one stereocentre and forms as equal amounts of both enantiomers (racemic, optically inactive). A meso form is not possible with one stereocentre.