Answer
$5.1\leq \sigma^2\leq 18.3$
Work Step by Step
Given $n=15, s^2=8.6, c=0.9$
use the $\chi^2$ table with $df=14, \alpha/2=0.05, 0.95$
we find $\chi^2_{right}=23.685$ and $\chi^2_{left}=6.571$
We can build the confidence interval as $\frac{14\times8.6}{23.685}$ to $\frac{14\times8.6}{6.571}$
which means the 90% confidence interval of the true variance is $5.1\leq \sigma^2\leq 18.3$
