The Question
Suppose I have the i.i.d. random variables $X_1,\dots,X_n \sim N(\mu,\sigma^2)$ where $\mu$ is known and $\sigma^2$ is not. Find a possible value for $a_1$ and $a_2$ such that the interval
$$\biggl (\sum_{i=1}^n \frac{(X_i- \bar X)^2}{a_1} \;,\; \sum_{i=1}^n \frac{(X_i- \bar X)^2}{a_2} \biggl)$$
is a 90% confidence interval for $\sigma^2$ in the case when $n=10$.
My Attempt
So, I guess that means we want to find $a_1$ and $a_2$ such that
$$\Bbb P \biggl (\sum_{i=1}^{10} \frac{(X_i- \bar X)^2}{a_1}<\sigma^2< \sum_{i=1}^{10} \frac{(X_i- \bar X)^2}{a_2} \biggl)=0.90$$
Rearranging, we have
$$ \Bbb P \biggl (a_2\sigma^2<\sum_{i=1}^{10} (X_i- \bar X)^2<a_1\sigma^2 \biggl)=0.90$$
And I know that
$$\sum_{i=1}^{10} (X_i- \bar X)^2 = \sum_{i=1}^{10} X_i^2-10\bar X^2$$
But where do I go from here? I am completely stuck. Any help would be much appreciated. Thanks.
Population mean known. If $\mu$ is known, then there is no need to estimate it by $\bar X.$ In that case $$Q=\frac{\sum_i (X_i - \mu)^2}{\sigma^2} = \sum_{i=1}^n \left(\frac{X_i -\mu}{\sigma}\right)^2 = \sum_{i=1}^n Z_i^2\sim \mathsf{Chisq}(df = n),$$ where $Z_i$ are independently standard normal. You can use software or tables to find numbers $L$ and $U$ with $P(L \le Q \le U) = .90.$ With a little algebra this gives a 90% CI for $\sigma^2.$
Population parameters unknown. In the case most often occurring in practice, neither $\mu$ nor $\sigma^2$ is not known. Then $\mu$ is estimated by $\hat \mu = \bar X.$ and $\sigma^2$ by $\widehat{\sigma^2} = S^2 = \frac{1}{n-1}\sum_i (X_i - \bar X)^2.$
In this case, one can show (but not as easily as above) that $$Q^\prime = \frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq(n-1)}.$$ Then $L$ and $U$ can be found such that $$.90 = P(L \le Q^\prime < U) = \cdots = P\left(\frac{(n-1)S^2}{U}\le \sigma^2 \le \frac{(n-1)S^2}{L}\right).$$ Then a 90% CI for $\sigma^2$ is of the form $\left(\frac{(n-1)S^2}{U},\,\frac{(n-1)S^2}{L}\right).$
For example, if $n = 10,$ you can use $L = 3.325$ and $U = 16.92$ from a printed table of the chi-squared distribution or from software. Results for R statistical software are shown below. That choice would give you a 'probability-symmetric' CI. By cutting slightly different probabilties from the tails (still adding to 10%), you could get a slightly shorter interval.
Below is a histogram of a million values of $Q^\prime$ based on samples of size $n = 10.$ The solid blue density curve for $\mathsf{Chisq}(9)$ fits well; the dotted curve for $\mathsf{Chisq}(10)$ does not. This is only a demonstration for a particular case.
A formal proof of the distribution of $Q^\prime$ can be done using linear algebra or moment generating functions. (As I recall, the proof relies on the fact that for normal data, $\bar X$ and $S^2$ are stochastically independent random variables. The proof is easy for $n = 2$ because $\bar X$ is a function of $X_1+X_2$ and $S^2$ is a function of $X_1 - X_2,$ where $X_1 + X_2$ and $X_1 - X_2$ are uncorrelated and thus, by normality, independent.)
Note: You say that $\mu$ is known and that you are supposed to make a CI using $\bar X.$ This makes no sense in practice. Please check. If you know $\mu$ and use $\bar X$ anyhow (as a textbook exercise), then you would have to use $\mathsf{Chisq}(n-1)$ as in the second part of my answer. A CI with $n-1$ DF will be a little longer than a corresponding one with $n$ DF. But if you include $\bar X,$ then you include its variability also, and your CI will not be optimal.