Answer
240 exams
Work Step by Step
Given $\sigma^2=900$, we have $\sigma=30$.
At the 99% confidence, the critical z-value is $z_{\alpha/2}=2.58$
Since $E=5$. we have the equation $E=2.58\times\frac{30}{\sqrt n}=5$
Thus, the required sample size is $n=(\frac{2.58\times30}{5})^2=239.6\approx240$ exams