I'll first put on here the null and alternative hypotheses. Because the student believes the mean time to be less than 5, this is your alternative.
H0: µ=5
H1: µ<5
Now for your test statistic, since we don't have the population standard deviation, we'll be getting a t test statistic (not z). Since there are 68 students surveyed, we have 67 degrees of freedom So we can set the test statistic up as follows:
t = (4.5 - 5)/(1.8/√(68)) = -2.29
Because the question asks for p-value, we can find this using a calculator. This would either be T.DIST(-2.29, 67) in Excel or tcdf(lower=-100, upper=-2.29, df=67) on a TI-84. In either case, this will get you a p-value of 0.0126. Since this is less than α=0.05, you'll be rejecting the null hypothesis and concluding that there is sufficient evidence to support the student's claim of fewer than 5 years to complete their degrees.