To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.
Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

The area of a sphere is given by

So replacing we have to

Since the question tells us to find the proportion when

So considering the two intensities we have to


The ratio between the two intensities would be

The power does not change therefore it remains constant, which allows summarizing the expression to

Re-arrange to find 



Therefore the intensity at five times this distance from the source is 