Given a 95% confidence level, margin of error E=0.1, and preliminary sample with standard deviation s = 2, \(z_{\alpha / 2} = 1.96\) gives
\begin{equation*}
n \gt \left ( 1.96 \cdot \frac{2}{0.1} \right )^2 \approx 1536.64
\end{equation*}
or a sample size of at least 1537.