The power of the test is probability of Type II error. IN this case it is probability of type II error when x-bar = 23.6. This is the probability of Ho not being rejected when it is false because x-bar = 23.6
B(23.6) = 1 - z(23.6 - 25.1)(14/sqrt(100))
1- z(-1.07)
Finr z(-1.07) on standard normal distribution chart to get .1423
1- .1423 = .8577
b) we know the power of the test is .9
so 1 - z(23.6 - 25.1)/(14/sqrt(n)) = .9
therefore z(-1.5/914/sqrt(n)) = .1
looking at the z-chart we know z(-1.28) = .1
so -1.28 = -1.5(14/sqrt(n))
-17.92/sqrt(n) = -1.5
cross multiply and square both sides and solve for n and you get n = 143.