25 uW by Rm
100 uW by RL
Are you a student?
There are many solutions:
1) use Spice, the most complicated.
2) use ADS or other freq. domain simulator and "normalize" the exiting voltage to 1.
3) use the brain as follow:
3a) "move" to left RL across the 2nd lamda/4 line... RL became R1=50^2/100=25 Ohm. R1 is shunted to Rm.
3b) calculate the parallel Rm//R1. Rm//R1=R2=100//25=20 Ohm.
3c) move to left R2. R3=50^2/R2 = 2500/20 = 125 Ohm
3d) since le lines are loss free, they cannot dissipate any power, so for energy conservation law, the power dissipate into R3 is the sum of the power dissipate into Rm and RL
3e) find power dissipate into R3
P3=(0.2/(75+125))^2*125=125 uW
3f) find Voltage applied to Rm.
Vm=(20*125uW)^.5= 50mV
3g) find power dissipated by Rm: Pm=0.05^2/100=25uW
3h) find power dissipated by RL: 125uW-25uW=100 uW