Answer: B
Moles of alkene =0.140/70=2.00 mmol and of ether =0.812/116=7.00 mmol. The substrate consumed is 2.00+7.00=9.00 mmol. The elimination percentage is 2.00/9.00×100=22.2%. (Dividing by the 10.0 mmol originally taken gives 20.0%, which is wrong because some bromide was not consumed.)