1. **State the problem:** We want to find the sample size $n$ required to estimate a population proportion $p$ within a margin of error $E=0.06$ at a 99% confidence level. 2. **Formula used:** The margin of error for a population proportion is given by $$E = z_{\alpha/2} \times \sqrt{\frac{p(1-p)}{n}}$$ where $z_{\alpha/2}$ is the critical value from the standard normal distribution for the desired confidence level. 3. **Find $z_{\alpha/2}$:** For 99% confidence, $\alpha=0.01$, so $z_{\alpha/2} = z_{0.005} \approx 2.576$. 4. **Rearrange formula to solve for $n$:** $$n = \left(\frac{z_{\alpha/2}}{E}\right)^2 p(1-p)$$ 5. **Substitute values:** $$n = \left(\frac{2.576}{0.06}\right)^2 \times 0.55 \times (1-0.55)$$ 6. **Calculate step-by-step:** - Calculate $\frac{2.576}{0.06} = 42.9333$ - Square it: $42.9333^2 = 1843.9$ - Calculate $p(1-p) = 0.55 \times 0.45 = 0.2475$ - Multiply: $1843.9 \times 0.2475 = 456.3$ 7. **Conclusion:** The required sample size is approximately $457$ (always round up to ensure the margin of error). **Final answer:** $\boxed{458}$