The problem is
Find the variance $S^2$ for random sample of size 21 from a normal population with variance 5.
(Hint: Use the fact that the statistic $\frac{(n-1)S^2}{\sigma^2}$ has a Chi-squared distribution with n-1 degrees of freedom, for a normal population with variance $\sigma^2$)
So I get that the statistic $\frac{(21-1)S^2}{5}$ has a Chi-squared distribution with 20 degrees of freedom. But I'm not sure how to proceed from there since I don't get how the Chi-squared distribution is related to getting the particular variance value.
Hint:
It is asked for $Var\left(S^2 \right)$. And we know that $\frac{(n-1)\cdot S^2}{\sigma^2} \sim \chi_{n-1}^2$ with $Var\left(\chi_{n-1}^2\right)=2(n-1)$. Thus
$$Var\left(\frac{(n-1)\cdot S^2}{\sigma^2} \right)=2\cdot (n-1),$$
where $(n-1)$ and $\sigma^2$ are constants.
$$\frac{(n-1)^2}{\sigma^4}Var\left( S^2 \right)=2\cdot (n-1)$$
Go on and solve the equation for $Var\left( S^2 \right)$.