Given the mean weight and population standard deviation;
[tex]\begin{gathered} n=32 \\ \sigma=4.7 \\ \bar{x}=53 \end{gathered}[/tex]The margin of error also called the maximum error of the estimate is given as;
[tex]E=Z_{\frac{\alpha}{2}}\times\frac{\sigma}{\sqrt[]{n}}[/tex]The z-value of 99% confidence interval is 2.576.
So,
[tex]\begin{gathered} E=2.576\times\frac{4.7}{\sqrt[]{32}} \\ E=2.576\times0.8309 \\ E=2.1403 \\ E\cong2.14ounces\ldots\ldots\ldots\ldots..(two\text{ decimal places)} \end{gathered}[/tex]