The correct option is (d) 200mW
The explanation: 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)
⇨ Pavg-hlf = 2Pavg mono = 2*100mW=200mW