Answer
$\approx 150\;\rm m$
Work Step by Step
First, we need to draw the force diagram exerted on you at the bottom point of the dip, as seen below.
Now we need to apply newton's second law;
$$\sum F_r=F_n-mg=ma_r=m\dfrac{ v^2}{R}$$
where $R$ is the radius of the dip.
$$ F_n-mg= m\dfrac{ v^2}{R}$$
We know that your own weight is 588 N, as mentioned by the author. Hence, your mass is given by $m=\dfrac{588}{9.8}=60\;\rm kg$.
Thus,
$$ \boxed{F_n-588 = \dfrac{60 }{R}v^2}$$
We need to use this formula to find the relationship between the speed squared and $(F_n-588 )$ which had to be a straight line since $R/60$ is constant.
The slope of this line will be equal to
$${\rm Slope}=\dfrac{60}{R}$$
Hence,
$$R=\dfrac{60}{{\rm Slope}}\tag 1$$
Noting that $y=F_n-588$ and $x=v^2$
Plugging the data into a table, as seen below.
\begin{array}{|c|c|c|}
\hline
x=v^2\;({\rm m^2/s^2})& y=F_n-588 \;({\rm N}) \\
\hline
(0^2=\color{blue}{\bf 0}) & (588-588=\color{blue}{\bf 0} ) \\
\hline
(5^2=\color{blue}{\bf 25}) & (599-588=\color{blue}{\bf 11}) \\
\hline
(10^2=\color{blue}{\bf 100}) & (625-588=\color{blue}{\bf 37}) \\
\hline
(15^2=\color{blue}{\bf 225}) & (599-588=\color{blue}{\bf 86}) \\
\hline
(20^2=\color{blue}{\bf 400}) & (756-588=\color{blue}{\bf 168}) \\
\hline
(25^2=\color{blue}{\bf 625}) & (834-588=\color{blue}{\bf 246}) \\
\hline
\end{array}
Plug these dots into a graph and connect the dots by the best-fit straight line and find its slope. Then plug the result into (1).
$$R=\dfrac{60}{{\rm Slope}}=\dfrac{60}{0.39}=153.4\;\rm m$$
Therefore,
$$R\approx \color{red}{\bf 150}\;\rm m$$
