Comment: You should be able to find formulas matching those below,
supported by suitable derivations, in an elementary textbook or online.
You don't say what kind of technology you might be expected to use, but
many statistical software packages have procedures that compute
confidence intervals for population variances.
Assuming your $n=25$ observations are a random sample from a normal population, with sample variance $S^2$ (denoted as v
in the R code below) estimating population variance $\sigma^2,$ you have
$$Q = \frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\text{df}=n-1).$$
Thus $$0.95 = P\{L \le Q \le U\} = P\left\{\frac{(n-1)S^2}{U} \le
\sigma^2 \le \frac{(n-1)S^2}{L} \right\},$$
where $L$ and $U$ cut probability 0.025 from the lower and upper tails of
$\mathsf{Chisq}(n-1),$ respectively.
Then a 95% confidence interval for $\sigma^2$ is of the form
$$\left(\frac{(n-1)S^2}{U}, \frac{(n-1)S^2}{L} \right).$$
For your data, a 95% CI is $(5.443, 17.277)$, computed in R is as follows:
x = c(91.46, 92.68, 89.54, 91.37, 82.87, 94.74, 89.27, 89.68, 84.25,
87.67, 89.33, 91.52, 84.64, 89.16, 92.92, 92.45, 84.45, 90.74,
92.36, 89.75, 89.37, 89.56, 92.68, 88.46, 89.53)
v = var(x); n = length(x)
[1] 8.927358 # sample variance
[1] 25 # sample size
(n-1)*v/qchisq(c(.975,.025), n-1)
5.442947 17.277155
Minitab software provides the following output (based on sample variance and sample size) along with a notice that the chi-squared method is
valid only for normal data.
N StDev Variance
25 2.99 8.93
95% Confidence Intervals
CI for CI for
Method StDev Variance
Chi-Square (2.33, 4.16) (5.44, 17.28)
Note: A Shapiro-Wilk test of normality indicates (with P-value about 5%)
that your data may not be normal. Notice that there are a few outliers at the
lower end of the sorted data.

If you have serious doubts about the normality of your data and your 'technology' includes bootstrapping, you might try that. I got
the 95% nonparametric bootstrap CI $(5.92, 20.44).$ Bootstrapping is a simulation procedure, so additional runs (with different seeds) will give
slightly varying results. Also, there are several different styles
of bootstrap confidence intervals, which might give somewhat different results.
set.seed(2019)
r = replicate(10^5, var(sample(x, n, repl=T))/v)
v/quantile(r, c(.975,.025))
97.5% 2.5%
5.923514 20.438291
For some further information on bootstraps see this page.
[self-study]
tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ – gung - Reinstate Monica Mar 4 '19 at 3:35