You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately $21 \%$. You would like to be $90 \%$ confident that your estimate is within $2.5 \%$ of the true population proportion. How large of a sample size is required? Do not round midcalculation. \[ n= \]

Step 1 ：We are given that the estimated proportion P is 0.21, the desired margin of error E is 0.025, and the z-score Z corresponding to a 90% confidence level is 1.645.

Step 2 ：We can use these values in the formula for sample size estimation for proportions, which is \(n = \frac{{Z^2 \cdot P \cdot (1-P)}}{{E^2}}\).

Step 3 ：Substituting the given values into the formula, we get \(n = \frac{{(1.645)^2 \cdot 0.21 \cdot (1-0.21)}}{{(0.025)^2}}\).

Step 4 ：Solving this expression, we find that the required sample size is 719.

Step 5 ：\(\boxed{719}\) is the final answer.