$Given: Downlink Frequency(f)=2.2GHz$
$Equivalent noise temperature(T_e )=160K$
$Bandwidth(B)=1MHz$
$Gain of the receiver [G_RX]=30dB$
$Antenna noise temperature[T_ant]=190dB$
$[\frac{C}{N}]_D=20dB$
Geostationary satellite distance $(r)=42164Km$
To find: $[EIRP]:$
$System Noise temperature[T_S ]=[T_ant ]+10logT_e$
$=190+10log160$
$[T_S]=212.041dB$
Considering Losses as only frequency dependent loss i.e. $FSL$
$[FSL]=10log(\frac{4πr}{λ})^2=10log(\frac{4πrf}{c})^2=10log(\frac{4π×42164×10^3×2.2×10^9}{3×10^8 })^2
[FSL]=191.789dB$
We Know
$[\frac{C}{N}]_D=[EIRP]+[\frac{G}{T_S}]_ES-10logk-10logB-[Losses]$
$[\frac{C}{N}]_D=[EIRP]+[G]_ES-[T_S]-10logk-10logB-[Losses]$
$20=[EIRP]+30-212.041-10 log(1.38×10^{-23} )-10 log(1×10^6 )-191.789$
$20=[EIRP]-205.23$
$[EIRP]=225.23dBW$