It's easiest to compare each of them

Observe that in all the populations form an arithmetic progression . This means that the middle element is the mean.

To have a large variance is for the elements to be scattered , i.e, non-mean elements to have a large deviation from the mean. In the context of an Arithmetic Progression, this means that the common difference should be as large as possible.

Clearly, {2, 4, 6} satisfies that criteria.

Using the formula for population variance on a data set of $n$ elements:

$\sigma^{2} = \Sigma_{i=1}^{n} \frac{(x_{i}-\mu)^2}{n}$

(where $\mu$ is the average (mean) of the elements), we find that:

$\sigma^{2}_{2,4,6}=\frac{8}{3}$

$\sigma^{2}_{2,3,4} = \sigma^{2}_{3,4,5} = \sigma^{2}_{1,2,3} = \frac{2}{3}$

The set $\boxed{2,4,6}$ yields the largest population variance.