Answer
$28.6\leq\sigma^2\leq 334.2$ and $5.3\leq\sigma\leq 18.3$
Work Step by Step
From the data set we have $n=11, s^2=72.05$
for $c=0.99$ use the $\chi^2$ table with $df=10, \alpha/2=0.005,0.995$
we find $\chi^2_{right}=25.188$ and $\chi^2_{left}=2.156$
we can then build a variance confidence interval as $\frac{10\times72.05}{25.188}$ to $\frac{10\times72.05}{2.156}$
this mean that the population variance is in the range of $28.6\leq\sigma^2\leq 334.2$
and the population standard deviation is in the range of $5.3\leq\sigma\leq 18.3$
