Right answer is (c) 236.5μW

To elaborate: given d=20m, Pt=50W and f=900MHz

Gain of half-wave dipoles is 1.64

\(λ = \frac{c}{f} = \frac{3×10^8}{900Mhz} = \frac{1}{3} m \)

\(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{(4\pi R)^2} = \frac{1.64×1.64×1/3^2}{(4\pi ×20)^2}\)

Pr=236.5μW