Correct answer is (a) 2sin^-1(λ/b)
To explain I would say: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. \(\frac{kb}{2}\) sinθ=nπ
θ= sin^-1\((\frac{n\lambda}{b})\)
Therefore, the FNBW in E-plane is given by FNBW=2 θ = sin^-1\((\frac{\lambda}{b})\).