Correct answer is (a) 50mW
To elaborate: Average Power radiated from the half-wave dipole \(P_{avg-hlf}=I_{rms}^2 R_{hlf}\)
⇨ \(\frac{P_{avg-hlf}}{P_{avg\, mono}} = \frac{R_{hlf}}{R_{mono}} = \frac{73}{36.5}=2\) (under same current)
\(P_{avg \,mono}=\frac{P_{avg-hlf}}{2}=\frac{100mW}{2}=50mW.\)