First note that $\mathbb{E}\left[(X-\mu)^2 \right] = \sigma^2$
An unbiased estimator of $\sigma^2$ is given by$\hat{s}^2 = \displaystyle \frac{\displaystyle \sum_{k=1}^{n} \left( X_i - \mu \right)^2}{n}$ Now we have that $\displaystyle \frac{n \hat{s}^2}{\sigma^2}$ is a $\chi^2_n$ i.e. a $\chi^2$ random variable with $n$ degrees of freedom (since $X_i$ are normally distributed random variables)
Find $a$ and $b$ such that $\displaystyle 1- \frac{\alpha}{2} = P\left(\chi_n^2 \leq b \right)$ and $\displaystyle \frac{\alpha}{2} = P\left(\chi_n^2 \leq a \right)$
Hence, we have $\displaystyle P \left(a \leq \chi_n^2 \leq b \right) = 1 - \alpha$ $\displaystyle P \left(a \leq \frac{n \hat{s}^2}{\sigma^2} \leq b \right) = 1 - \alpha$ $\displaystyle P \left(\frac{n \hat{s}^2}{b} \leq \sigma^2 \leq \frac{n \hat{s}^2}{a} \right) = 1 - \alpha$
Hence, the desired $1-\alpha$ confidence interval is given by $\left(\frac{n \hat{s}^2}{b},\frac{n \hat{s}^2}{a} \right)$ where $a$ and $b$ are given by the $\displaystyle F_{\chi}(a) = \frac{\alpha}{2}$ and $\displaystyle F_{\chi}(b) = 1 - \frac{\alpha}{2}$ where $F_{\chi}$ denotes the cumulative distribution function.