Solution:

z = 30 km

P(0) = 200 µW

Attenuation in optical fiber is given by,

$\begin{aligned} 0.8=10 \times \frac{1}{30} \log \left[\frac{200 \mu \mathrm{W}}{\mathrm{P}(\mathrm{z})}\right] \\\\ 2.4=10 \times \log \left[\frac{200 \mu \mathrm{W}}{\mathrm{P}(\mathrm{z})}\right] \end{aligned}$