2 in 1000 gives a mean of 0.2%=0.002. In 20000 children µ, the mean, is 0.002*20000=40. So we would expect 40 cases in a sample size of 20,000.
a) Assuming a binomial distribution, in the absence of any other statistic, SD=√(20000*0.002*0.998)=6.32 approx. If we use this as σ in a normal distribution the z-score for X=25 is (25-40)/6.32=-2.37 corresponding to 0.0089 or 0.89%.
b) It is unlikely for X=25 given the mean and SD calculated.