Yes you are right. You have $$df = n - 1 = 21$$ degrees of freedom for the $t-$distribution. Here we have a 2-sided test, so we should be looking at a right (or left) tailed area of $\frac{\alpha}{2} = \frac{0.02}{2}=0.01$ with $21$ degrees of freedom. Using the table below, we can extract the critical value for the desired confidence interval. (At the rows, you should be looking at $df = 21$ and at the columns you should be reading $0.01 $).

The key to this problem is that they say you are working with $s^2$, which is an estimate of the population variance. You have applied a z-distribution to the problem, which would be correct if you had the actual population variance ($\sigma^2$) . However, when you are working with an estimate of the population variance you must apply a student's t-distribution to the problem.

The form of the t-distribution equation is similar to the z-distribution equation except that you replace the "z" with a "t", and the population standard deviation with the sample standard deviation (s):

$t = \frac{xbar - \mu}{\frac{s}{\sqrt{n}}}$

where $s$ is the square root of the sample variance, $n$ is the sample size, and $t$ is a value to be calculated from a t-table. You will need two things to find the t-value to plug into your equation:

1. Confidence interval - in this case you are asked for 99% confidence, meaning that your total acceptable Type I error rate ($\alpha$) is 1-.99 = .01. That error rate must be split over both ends of the t-distribution, thus the heading of the t - table column you seek should read .005.
2. Degrees of freedom ($n$). In this case, your Degrees of Freedom will be 5-1 = 4

Go to a t-table with a t($\alpha$/2=.005, DOF=4), and you will have enough information to find your answer.