Most of the information to solve this problem is provided in the statement, therefore we will apply the concepts related to the intensity of the sound and its method of representation across the logarithmic scale.
By definition as we saw the level of sound intensity in decibels is represented by
[tex]\beta = 10log(\frac{I}{I_0})dB[/tex]
Where, I = Intensity for which decibels is to be calculated
[tex]I_0[/tex]= Reference intensity (at this case is [tex]10^{-12}W/m^2[/tex]
PART A ) Intensity is 10 times the reference intensity.
Here [tex]I = 10I_0[/tex], replacing
[tex]\beta = 10log(\frac{10I_0}{I_0})dB[/tex]
[tex]\beta = 10log(10)dB[/tex]
[tex]\beta = 10dB[/tex]
Therefore the sound intensity in decibels of a sound wave 10 times stronger than reference intensity is 10dB
PART B) Intensity is 100 times the reference intensity
Here [tex]I = 100I_0[/tex], replacing
[tex]\beta = 10log(\frac{100I_0}{I_0})dB[/tex]
[tex]\beta = 10log(100)dB[/tex]
[tex]\beta = 20dB[/tex]
Therefore the sound intensity in decibels of a sound wave 10 times stronger than reference intensity is 20dB