The best way to determine whether 5 is an unusual number to have for this blood type out of a sample size of 45 people is to determine how many standard deviations 5 is from the mean. As IATM points out, the mean is 2.7. The formula for standard deviation of a binomial distribution like this is:
where for this problem n = 45 and p = 0.06. So here the variance is 45(/06)(.94)= 2.54, and the standard deviation is

. So having 5 people with this blood type is (5-2.7)/1.6 = 1.44 standard deviations from the mean. So now you need to decide whether a Z-score of 1.44 is "unusual" or not. Hope this helps.