Answer
See the detailed answer below.
Work Step by Step
We are given that
$$I=\dfrac{cP_S}{r^x}$$
and we are given data of $r$ and $\beta$ and we need to find $x$ by drawing the best fit line. So we have to do some math.
Let's take the $\log$ for both sides;
$$\log_{10} I=\log_{10}\left[\dfrac{cP_S}{r^x}\right]$$
$$\log_{10} I=\log_{10} (cP_S)- \log_{10} (r^x)$$
$$\log_{10} I=\log_{10} (cP_S)-x \log_{10} (r)\tag 1$$
Recalling that
$$\beta=10\log_{10}\left(\dfrac{I}{I_0}\right)$$
Hence,
$$\beta=10[\log_{10} (I)-\log_{10}(I_0)]$$
Plugging from (1);
$$\beta=10[\log_{10} (cP_S)-x \log_{10} (r)-\log_{10}(I_0)]$$
where $I_0=1\times 10^{-12}$
$$\beta=10[\log_{10} (cP_S)-x \log_{10} (r)-\log_{10}(1\times 10^{-12})]$$
$$\beta=10[\log_{10} (cP_S)-x \log_{10} (r)+12]$$
$$0.1\beta= \log_{10} (cP_S)-x \log_{10} (r)+12 $$
Now this is an equation of a straight line $y=mx+b$ where $y=0.1\beta$,
$x=\log_{10}(r)$,
$m={\rm slope}=- x$,
and
$b=\log_{10} (cP_S)+12$
$$0.1\beta= -[\log_{10} (r)]x +[\log_{10} (cP_S)+12 ]$$
Use the given table to find the following table;
\begin{array}{|c|c|c|c|}
\hline
x=\log_{10} (r)&y =0.1\beta \\
\hline
\log_{10} (1)=0& 10 \\
\hline
\log_{10} (3) & 9.3 \\
\hline
\log_{10} (10)=1& 8.5 \\
\hline
\log_{10} (30)& 7.8 \\
\hline
\log_{10} (100)& 7 \\
\hline \end{array}
And as we see in the graph below, the slope of the best fit line is $-1.5$ and hence,
$$x=\color{red}{\bf 1.5}$$
