Using $F$-distribution
If $s^2_1$ and $s^2_2$ are the variances of independent samples of sizes $n_1$ and $n_2$, respectively, from normal populations, then a $100(1-\alpha)\%$ C.I. for $\sigma^2_1 / \sigma^2_2$ is
$$ \frac{S^2_1}{S^2_2}\frac{1}{f_{\alpha / 2}(n_1-1, n_2-1)} \lt \frac{\sigma^2_1}{\sigma^2_2} \lt \frac{S^2_1}{S^2_2}f_{\alpha / 2}(n_2-1, n_1-1) $$
where $f_{\alpha/ 2}(v_1,v_2)$ is an $f$-value with $v_1$ and $v_2$ df, leaving an area of $\alpha / 2$ to the right, and $f_{\alpha / 2}(v_2,v_2)$ is a similar $f$-value with $v_2$ and $v_1$ df.
Example 7 gives
$\bar x_1 = 3.84, \bar x_2 = 1.49$
$s_1 = 3.07, s_2 = 0.80$
$n_1 = 15, n_2 = 12$
Task: find $\sigma^2_1 / \sigma^2_2$ for 98% C.I.