The correct choice is (a) 4.36°
For explanation: Given NACA 22104 airfoil.
AOA = 3°, Cl = 0.15 at 3°
Based on NACA -5 digit series design Cl = first digit*0.15 = 2*0.15 = 0.3
Now, for thin airfoil lift curve slope = 0.11 per degree
0.11 = (design Cl – Cl at 3°) / (AOA –3°)
0.11 = (0.3-0.15) / (AOA -3°)
0.11*(AOA -3°) = 0.15
Hence, at designed Cl, AOA = (0.15/0.11) + 3° = 4.36°.