Assuming there are a large enough sample such that each selection is independent (i.e. choosing a standard item does not affect the probability that the next selection is a standard item) you can model a problem like this using a binomial distribution.
The random variable $X$ (the number of standard type items selected) is distributed binomially with $n=2$ (two selections) and $p=0.8$ (probability of success, i.e. choosing an item of standard type with each selection is $0.8$). That is: $X\sim B(2,0.8)$.
Therefore, using the P.M.F for the binomial distribution, we have:
Which is the answer that you have obtained, so the book is indeed mistaken.