$\beta=10 \log _{10}\left(\frac{I}{I_{0}}\right)$

where, $I=$ intensity of sound and $I_0=$ reference intensity $\left(\sim 10^{-12} \,W / m ^2\right.$ ).

Now, taking antilog, we have

$\frac{I}{I_0}=10^{\left(\frac{\beta}{10}\right)} \Rightarrow I=I_0 \times 10^{\left(\frac{\beta}{10}\right)} \quad \dots(i)$

Here, $\beta=20$ at $1 \,kHz$ and $\beta=60$ at $9 \,kHz$.

'Taking a linear relation, $\beta=k f+c$

where, $k$ and $c$ are constants and $f=$ frequency.

So, we have

$20=k \times 1+c \quad \dots(ii)$

$\text { and }$ $60=k \times 9+c \quad \dots(iii)$

Subtracting, Eqs. $(ii)$ and $(iii)$, we get

$60-20 =9 k-k+c-c$

$40 =8 k+0$

$8 k=40$

$k=5$

Putting the value of $k$ in Eq. $(ii)$, we have

$20=5+c \Rightarrow c=15$

$\Rightarrow \quad c=15$

$\therefore \operatorname{At} f=5 \,kHz ,$

$\beta=k f+c=5(5)+15$

$=40$

Now, width $\beta=20$ and $\beta=40$ from Eq. $(i)$, we have

$\text { Intensity, } \quad I =I_{0} \cdot 10^{\left(\frac{\beta}{10}\right)}$

$\Rightarrow \quad I_{1 \,kHz }=I_0(10)^{\frac{20}{10}}=I_{0} \cdot 10^2$

$I_{5 \,kHz }=I_0(10)^{\frac{40}{10}}=I_0 \cdot 10^4$

$\Rightarrow \quad \frac{I_{5 \,kHz }}{I_{1 \,kHz }}=\frac{10^4}{10^2}=100$