Let $\hat{p}$ be the sample proportion of Republican voters, and let $p$ be the true proportion of Republican voters. You would like $P(|\hat{p}-p| \le 0.02)$ to be $90\%$.
The mean of the random variable $\hat{p}$ is $p$, and the standard deviation of $\hat{p}$ is $\sqrt{\frac{p(1-p)}{n}}$, where $n$ is the sample size.
A not unreasonable estimate for $p$ is that it is about $0.42$. So a not unreasonable estimate for the standard deviation of $\hat{p}$ is $\sqrt{\frac{(0.42)(0.58)}{n}}$.
By the normal approximation to the binomial, we want $1.645$ "standard deviation units" to be about $0.02$. In symbols, $1.645 \sqrt{\frac{(0.42)(0.58)}{n}}\approx 0.02.$ Now you can use some algebra to solve for $n$. You may have been supplied a ready-rolled formula for $n$. But you can also use the information here to roll your own.
Comment: If one expects a probability to be not very far from $0.5$, the unknown term $\sqrt{p(1-p)}$ can be assumed to be about $0.5$. Because of our prior knowledge about the rough value of $p$, we used $\sqrt{(0.42)(0.58)}$ instead. This really makes very little difference, since $\sqrt{(0.42)(0.58)}$ is about $0.4936$, which is awfully close to $0.5$.
It would be sensible not to use the $0.42$ information at all, but I imagine you are expected to use it. Please note that if $p$ is very far from $0.5$, then $\sqrt{p(1-p)}$ will not be close to $0.5$.