Answer
$6.55\;\rm N/m$
Work Step by Step
We know that the peroid of one oscillation is given by
$$T=2\pi \sqrt{\frac{m}{k}}$$
So,
$$T^2=\left(\dfrac{4\pi^2}{k}\right)m\tag 1$$
We can use this formula to find the spring constant $k$ by considering $y=T^2$, $x=m$, and the slope of this line is given by
$${\rm Slope}=\dfrac{4\pi^2}{k}$$
Solving for $k$;
$$k=\dfrac{4\pi^2}{{\rm Slope}}\tag 2$$
Now we need to draw the best fit line of formula (1) by using the data in the table below.
Noting that the period is one-tenth the given time in the given table, as the author told us.
\begin{array}{|c|c|}
\hline
m\;{(\rm kg)}& T^2 \;{(\rm s^2)} \\
\hline
0.100 & 0.78^2 \\
\hline
0.150 & 0.98^2\\
\hline
0.200 & 1.09^2\\
\hline
0.250 & 1.24^2 \\
\hline
\end{array}
As we see in the graph below, the slope is about 6.03 s$^2$/kg. Plug that into (2).
$$k=\dfrac{4\pi^2}{6.03 }=\color{red}{\bf 6.55}\;\rm N/m$$
