The correct answer is (a) 5.71m
The explanation: Given location of spar = 65% of MAC, distance from t.e. = 2m.
Spar is located at 65% of chord from l.e. hence, from t.e. location of spar = (100% – 65%) of chord = 1-0.65 = 0.35*chord.
From given, 0.35*chord = 2m
Chord = 2/0.35 = 5.71m.