The time of execution for one program follows a normal distribution. For a sample of size $40$ we have: $\overline{x} = 32.2 s$ and $\sigma^{2} = 3.1s^{2}$
How many times we have to execute the program to obtain a $95\%$ confidence interval with the lenght of interval less than 2 seconds?
Because the program follows a normal distribution we have that de confidence interval is
$$\left[ \frac{1}{n}\sum X_{i} - \frac{Z_{\alpha/2}\sigma}{\sqrt{n}},\frac{1}{n}\sum X_{i} + \frac{Z_{\alpha/2}\sigma}{\sqrt{n}} \right]$$
so, the lenght interval is $$2\frac{Z_{\alpha/2}\sigma}{\sqrt{n}}$$
and we need that $$\frac{Z_{\alpha/2}\sigma}{\sqrt{n}} < 1$$ I first thought that I could use $\sigma = \sqrt{3.1}$ but this was for the sample of size $40$ so i don't know if I can use it.
In your situation you know the distribution is Gaussian but you eve do not know its variance. Given a sample size =40 you can assume that your $\hat{\sigma^2}=3.1$ is a good approximation...thus your reasonig is good and you can find
$$n\ge 1.96^2\cdot3.1\approx 11.91=12$$