It looks like a normal distribution has been used to find c.
c represents the number of standard deviations above the mean your value should be so that the probability of being below that value is 0.995.
If you scan the normal table at the back of the book for 0.995, you'll find z is around 2.58.
Page 504 also has a table on the topic.
Ok I see it is normal because n >=30 (or 30). If it was less than 30 then I would have used t distribution.