Asound from a source has an intensity of 270 db when it is 1 m from the source. what is the intensity of the sound when it is 3 m from the source?

Sound intensity is inversely proportional to the square of the distance between the source and the receiver.
That is 
I = k/r^2
 where
 k = constant
 r = radius

When r=1,  the intensity is I₁ = k/1 = k
When r=3, the intensity I₂ = k/3² = k/9
Therefore
 I₂ = I₁ /9

In decibels,
I = 10 log₁₀(I/I₀)
where I₀ = reference intensity

When r=1,
10 log₁₀ (I₁/I₀) = 270

When r =3,
10 log₁₀ (I₂/I₀) = 10 log₁₀ [(I₂/I₁)*(I₁/I₀)]
                     = 10 log₁₀ [(1/9)*(I₁/I₀)]
                     = 10 log₁₀(1/9) + 270
                     = 260.5

260.5 dB (nearest tenth)

No it doesnt because it can be any number.