Answer: A, B
E2 needs the leaving group and the beta-H anti-periplanar, i.e. both axial. In menthyl chloride Cl is equatorial in the stable chair, so the ring must flip to the all-axial form, which is high in energy; in that form the C2 isopropyl is axial, so the C2-H is equatorial and not anti to Cl, and only the C6-H is available (less substituted alkene). Neomenthyl chloride already has axial Cl in its stable chair with axial H on C2 and C6, so it eliminates faster.