Sound level, $L = 10 \, \log \left( \frac{I}{I_0}\right) $
Let intensity of 40 dB sound be $I$, then for 90 dB sound, intensity will be $I_x$.
$ \therefore \, 40 dB = \left(10 dB\right) \log\left(\frac{I}{I_{0}}\right)$ ....(i)
and $ 90 dB = \left(10 dB\right) \log\left(\frac{I_{x}}{I_{0}}\right) $.....(ii)
or, $ \log \left(\frac{I}{I_{0}}\right)= 4 $ and $ 9 = \log \left(\frac{I}{I_{0}}\right) + \log x $.
or ,$ 9= 4 + \log x or , \log x = 5$.
$\therefore \, x = 10^{5} $