Answer:
Given that
β= 120 dB at 5 m
We know that sound intensity given as


I= 1 W/m²
Power producing by loud speaker
P = I x 4πr²

P=314.15 W
a)
Let take intensity at 35 m is I'
r'=35 m
I'=P/4πr'²

I'=0.0204 W/m²
b)
Given that
β'= 80 dB


I'=10⁻⁴ W/m²
Let's take at distance r intensity is 80 dB
P=314.15 W
P = I' x 4πr'²
314.15 = 10⁻⁴ x 4πr'²
r'=500.11 m