Answer
(a) $83\ dB$.
(b) $31623 $ people.
Work Step by Step
Given $D=10log(\frac{I}{I_0})=10log(10^{12}I)=120+10log(I)$, we have:
(a) $D_1=80 \Longrightarrow 80=120+10log(I_1) \Longrightarrow 10log(I_1)=-40$. For $I_2=2I_1$, we have $D_2=120+10log(I_2)=120+10log(2I_1)=120+10log2+10log(I_1)=120+10log2-40\approx83\ dB$.
(b) Assume we need $x$ number of people for the purpose, we have $125=120+10log(xI_1) \Longrightarrow 10log(x)+10log(I_1)=5 \Longrightarrow 10log(x)=45 \Longrightarrow x=10^{4.5}\approx31623 $ people.
