Using the computational forumaula (or your calculator) gives \(s^2 \approx 479.5\text{.}\) Also, notice for n=9, the resulting interval will use a Chi-square variable with 8 degrees of freedom. Using the symmetric option, gives \(\chi_{0.025}^2 = 2.18\) and \(\chi_{0.975}^2 = 17.53\text{.}\) Therefore
\begin{equation*}
E_1 = \frac{8 \cdot 479.5}{17.53} \approx 221.095
\end{equation*}
and
\begin{equation*}
E_2 = \frac{8 \cdot 479.5}{2.18} \approx 1759.63.
\end{equation*}
Hence, you are 95% certain that
\begin{equation*}
221.095 \lt \sigma^2 \lt 1759.63.
\end{equation*}
By taking square roots you get
\begin{equation*}
14.87 \lt \sigma \lt 41.95.
\end{equation*}
Notice, this interval is relatively wide which is a result both of the number of data values being relatively small (n=9) and the actual data values being relatively large and spread out.