Answer :
To solve this problem we apply the concepts related to the Intensity, which is defined as the Power given by the object on unit area. Mathematically this is,
[tex]I = \frac{P}{A}[/tex]
Where,
I = Intensity
P = Power
A = Area
Considering that a whisper is around to [tex]1*10^{-10}W[/tex], then we have that
[tex]I = \frac{1*10^{-10}}{2^2}[/tex]
[tex]I = 2.5*10^{-11} W/m^2[/tex]
Intensity level in
[tex]\beta = 10log (\frac{I}{I_0})[/tex]
Where,
[tex]I_0 =[/tex] Threshold intensity level
[tex]\beta = 10log (\frac{2.5*10^{-11}}{1*10^{-12}})[/tex]
[tex]\beta = 14dB[/tex]
Therefore the corresponding sound intensity level is 14dB