Estimating p in the Binomial Distribution:

I have figured everything up to the point of how they are getting the answer of .0395 when asked to find a 99% confidence interval for p

E=(approx) Zc^sq.rt. Pcap (1-Pcap)/n=(approx) 2.58 ^sq. rt. (.75)(.25)/800 =(approx) .0395