Answer
$1.8\lt\sigma\lt3.2$
Work Step by Step
Given $n=20,s=2.3$, at a 90% confidence and $df=19$ the critical values are $\chi^2_{left}=10.117 $ and
$\chi^2_{right}=30.144 $, the range for the standard deviation can then be found as
$\sqrt {\frac{(n-1)s^2}{\chi^2_{right}}}\lt\sigma\lt\sqrt { \frac{(n-1)s^2}{\chi^2_{left}}}$
which gives $1.8\lt\sigma\lt3.2$
