Answer
$67.6\frac{W}{m^2}$
Work Step by Step
We know that the intensity level for sound 1 is:
$\beta_1=10log(\frac{I}{I_{\circ}})=10log(\frac{38.0}{10^{-12}})=135.80dB$
Now given that the intensity level of sound 2 is 2.5 dB greater than that of sound 1:
$\implies \beta_2=\beta_1+2.5dB=135.80+2.5=138.30dB$
Now we can find the intensity of sound 2 as:
$\beta_2=10log(\frac{I_2}{I_{\circ}})$
$\implies I_2=I_{\circ}10^{\frac{\beta_2}{10}}$
$I_2=(10^{-12})(10^{\frac{138.30}{10}})=67.6\frac{W}{m^2}$
