Answer
a. 51 dB.
b. $5\times10^{-9}W$.
Work Step by Step
a. Find the gain.
$$\beta = 10log \frac{P_{out}}{P_{in}}=10log \frac{135\;W}{1.0\times10^{-3}W}=51\;dB $$
b. Start with the gain equation, then solve for the noise power.
$$\beta = 10log \frac{P_{signal}}{P_{noise}}$$
$$ P_{noise}=\frac{P_{signal}}{10^{\beta/10}}=\frac{10\;W}{10^{93/10}}= 5\times10^{-9}W $$
