Verified answer

When the strength of a sound is given in dB the quantity being quoted is the intensity. dBs are a way of expressing the ratio of the intensity of a sound relative to a reference level of 10^-12W/m^2.

+80dB corresponds to an intensity ratio of 10^8, so a sound of 80dB has an intensity of 10^-4W/m^2.

In completely free air, a long way from any objects which might reflect the sound (for example if the source is very high in the air, and there is no wind), you can assume that the sound from a source spreads out uniformly over the surface of a sphere, so that if the power of the source is P, then the intensity at a radius r is P/(4*pi*r^2). If we assume that this question refers to such a situation we can write 10^-4 = P/(4*pi*20^2)

P = 0.50W

Note that in order to answer the question we have had to postulate a very unusual situation. If the sound source is in a more normal environment this calculation will be invalid. Sound intensity variations only follow an inverse-square law in idealised conditions quite unlike those normally encountered.