I found this part of a solution,
$P(p - 1.96 \frac{0.5}{\sqrt(n)} \leq \overline{X} \leq p + 1.96 \frac{0.5}{\sqrt(n)}) = 0.95$
$ \implies 1.96\frac{0.5}{\sqrt(n)} = 0.01$ (I don't get this implication part) Given that $\overline{X} = \frac{X_1 + ... + X_n}{n}$ with $X_i$ is a Bernouli distribution with $1$ if success and $0$ if failure, and $p = 0.5$.
Can you explain how we got to that implication?
You want to estimate the expectation $p$ via the mean value. The question now is how many single measurements have to enter the mean so that the expectation value is correct to about 2 digits after the dot with $95\%$ certainty, $$ P(|\bar X-p|\le 10^{-2})\le 0.95. $$ It is known that $P(|\bar X-p|\le 1.96·n^{-1/2}\sigma)=0.95$ if the single $X_k\sim N(p,\sigma)$. From your formula, $\sigma=0.5$. Thus one needs $$ 1.96·n^{-1/2}·0.5\le 10^{-2}\iff \sqrt{n}\ge 98. $$