The traditional (Wald) 95% CI for success probability $p$ uses the MLE $\hat p = x/n,$ where $x$ is the number of successes in $n$ trials. It is an asymptotic
CI intended for use with large $n$ where the normal approximation is accurate. It is of the form
$$\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}}.$$
In judging the $n$ required for a given margin of error
$E = 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$ it is customary to assume $\hat p = 1/2,$ which gives the maximum margin of error for a particular $n.$
The Agresti-Cooil CI uses $n^+ = n+4$ instead of $n$
and $p^+ = \frac{x+2}{n+4}$ instead of $\hat p.$ It comes closer to the intended 95% coverage of parameter $p$ for smaller values of $n.$
You can answer your questions about decreasing the width of the CI by a factor of 10, by deciding how
to decrease $E$ by a factor of 10.