Right answer is (b) λ/6
The explanation is: For an Hertzian dipole, equating the magnitudes of maximum induction and radiation fields we get,
\(\mid \overline{E_θ}\mid_{maxRadiation} =\mid \overline{E_θ}\mid_{maxInduction} \)
\(\frac{I_m dl}{4\pi\epsilon} (\frac{ω}{v^2r})=\frac{I_mdl}{4\pi\epsilon} (\frac{1}{r^2v})\)
\(r=\frac{v}{ω}=\frac{\lambda f}{2\pi f}=\frac{\lambda}{6}.\)