We want to estimate the proportion of customers each of whom uses the particluar brand of detergent.
$$\begin{align} \underbrace{p:=\text{population proportion which satisfies}~~0.8<p<0.9 }_{\text{known information} } \end{align}$$
And we want to determine the least number of samples for satisfying the following conditions.
$$\begin{align} P(\left| \hat p-p \right| \leq 0.02 )\geq 0.99 ~~\text{where}~~\hat p ~ ~\text{is a sample proportion} \end{align}$$
The following is a mix of the official answer(Only quoted partially) and my thoughts for the solution.
We assume $~ n\gg1 ~$
$$\begin{align} \hat p &= {x \over n }\\ x&:=\text{number of customers who uses the detergent from}~n~\text{samples} \\ X&\sim\mathcal B(n,p) \end{align}$$
Firstly, I can't understand why $~ p ~$ is used above instead of $~ \hat p ~$. Even I can't understand why binomial distribution can be applied to this random variable. Of course the $~ p,\hat p ~$represents a probability of randomly chosen person says that one uses a detergent of the brand, which can be handled as a boolean value hence intuitively the random variable can be of a binomial distribution.
From the assumption $~ n\gg1 ~$ ,
$~ X\sim\mathcal N(np,np(1-p)) ~$ can be satisfied approximately. BTW I define $~ q:=1-p ~$
$$ Z:= {X-np \over \sqrt{npq} } \sim\mathcal N(0,1) $$
$$ \underbrace{P(\left| Z \right|<2.576 )=0.99}_{\text{This can be derived using the table of std norm } } $$
$$ \underbrace{P \left(\left| \hat p-p \right| <2.576 \sqrt{{pq \over n }} \right)}_{\text{This can be derived using the above 2 eqns.} } =0.99 $$
The following is the main problem of this post.
$$ \underbrace{\color{fuchsia}{2.576 \sqrt{{pq \over n }}\leq 0.02 } }_{\text{How this can be derived?} } $$
Definitely the following 2 eqns are the clinchers I presume.
$$\begin{align} \underbrace{P(\left| \hat p-p \right| \leq 0.02 )\geq 0.99}_{\text{Constraint}}\\ \underbrace{P \left(\left| \hat p-p \right| <2.576 \sqrt{{pq \over n }} \right) =0.99 }_{\text{Known holdable eqn} } \end{align}$$
Intuitively I think that the following inequality should be satisfied which sign is opposite of the pink inequality.
$$ {\color{green}{2.576 \sqrt{{pq \over n }}\geq 0.02 } } $$
Let $F(\alpha) = P(|Z| \le \alpha)$. Then $F(\alpha)$ is strictly increasing.
We have $$P \left(\left| \hat p-p \right| \le 0.02 \right) = P\left(\left|Z\right| \le 0.02\sqrt{\frac{n}{pq}} \right) \ge 0.99.$$
Since $P(|Z| < 2.576) = 0.99$, we have $$0.02\sqrt{\frac{n}{pq}} \ge 2.576.$$