How many to ask

Idea. Guard against the worst case. The product $\hat{p}(1-\hat{p})$ is at most $\frac{1}{4}$, so size the sample for that value.

Step 1: worst case. $\hat{p}(1-\hat{p})\le \frac{1}{4}$, with equality at $\hat{p}=\frac{1}{2}$. So the half-width is at most $\frac{1.96}{2\sqrt{n}}$.

Step 2: solve. $$\frac{1.96}{2\sqrt{n}}\le 0.01⟺\sqrt{n}\ge 98⟺n\ge 9604.$$

The smallest such $n$ is $9604$.

Rule of thumb. Halving the margin of error requires four times as many observations, since the half-width scales like $\frac{1}{\sqrt{n}}$. A margin of $\pm 3$ points needs about $1100$ people, $\pm 1$ point about $9600$.