By what factor must the sound intensity be increased to increase the sound intensity level by 12.5 db ?

The sound has a unit of measurement. In physics, there are two measurements: in intensity (Watts per square meterI or in decibels. The most common unit of measurement is in decibels or dB. In fact, the threshold of hearing when it comes to measuring noise is in terms of decibels. 

To convert decibels to intensity in W/^2, the equation is

dB = 10log(I/I0), where I0 is equal to 10^-16. It is an empirical constant for the standard reference intensity. Therefore,

12.5 = 10 log (I/10^-16)
I = 1.78 × 10^-15 W/m^2

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way