Given a 99% confidence level, margin of error E=0.03, and preliminary estimate \(\tilde{p_0} = 0.35\text{,}\) notice that \(z_{\alpha / 2} = 2.58\) gives
\begin{equation*}
n \gt \left ( \frac{2.58}{0.03} \right )^2 0.35 \cdot 0.65 \approx 1682.59
\end{equation*}
or a sample size of at least 1683.