Let $n := |V| = rk$ be the total number of vertices in the graph. Basically, we are looking for a partition of $V$ into $r$ sets each of size $k$. The total cost will be then the sum of the weights of all edges present in the graph induced by each of these sets. Note that the edges having one end in one of these sets and the other endpoint in another set form together an $r$-cut and are exactly the edges we do not count in our sum. Hence, instead of minimizing the sum of edges in the subgraphs induced by the sets of the partition, we can maximize the weights of edges having one end in a set of the partition and the other end in a different set.
Note that we are looking here at a special variation of maximum $k$-cut, where we require that all the subsets of the partition have equal sizes and hence the problems are not completely equivalent. However, it is not hard to reduce the maximum $k$-cut problem (let us call it for now maximum $s$-cut to avoid confusion) to your problem, where given an instance of the $s$-cut problem (with $N$ vertices), you can add $(s-1)N$ additional vertices to the graph and connect each of them with all vertices in the graph with edges of weight 0. Set $k = N$ (note that $n = sN$) and hence, we will have $r = s$. Each solution to your problem corresponds to a solution to the maximum cut problem with dummy vertices assigned arbitrarily to make the count consistent.
The maximum cut problem is APX-hard and hence does not admit PTAS (polynomial time approximation scheme) unless P=NP. Using the previous reduction, and since the maximum cut problem is a special case of the maximum $k$-cut problem, your problem is at least as hard as the maximum cut problem.