Correct option is (a) 10.5m/s
The best I can explain: Given, civil aircraft, maximum CL = 3, wing loading W/S = 120
Approach speed of civil aircraft is given by,
Va = 1.3*Vstall
Where, Vstall = \(\sqrt{2*(\frac{W}{S})/(ρ*CL)}
= \sqrt{2*(120)/(1.225*3)}\) = 8.08m/s
Now, Va = 1.3*Vstall = 1.3*8.08 = 10.5m/s.