Answer
$4.34\times10^{7}\frac{dB}{watt/m^{2}}$
Work Step by Step
Loudness of a sound with intensity $I, L=10log_{10}(\frac{I}{I_{0}})$
Given: $L = 50$ dB
$10log_{10}(\frac{I}{I_{0}})=50$ dB
$log_{10}(\frac{I}{I_{0}})=5$ dB
or
$I=(10)^{5}{I_{0}}$ ...(1)
Using equation (1) and given value of intensity
$I=10^{-7}watt/m^{2}$
The rate of change of loudness is $=\frac{dL}{dI}$
Therefore,
$\frac{dL}{dI}=\frac{10}{ln10}\frac{1}{I}$
Hence,
$\frac{dL}{dI}=\frac{10^{8}}{ln10 } watt/m^{2}\approx 4.34\times10^{7}\frac{dB}{watt/m^{2}}$
