The expression for coupling factor is
\(C=10\log{\left(\frac{P_1}{P_3}\right)}=3\)
\(\log{\frac{P_1}{P_3}}=0.3\)
\(\frac{P_1}{P_3}=2\)
\(P_3=\frac{2.8\times{10}^{-3}}{2}=1.4\ mW\)
The expression of directivity is
\(D=10\log{\left(\frac{P_3}{P_4}\right)}=40\)
\(\frac{P_3}{P_4}=10000\)
\(P_4=\frac{1.4\times{10}^{-3}}{10000}=0.14\ µW\)
Thus the power at port 2 is
P2=Input power-(Power in coupling port+Power in isolated port)
P2=2.8×10-3-(1.4×10-3 + 0.14×10-6)=1.4 mW