= Solution
The <edge connectivity> $\lambda(G)$ is the minimum number of edges whose deletion disconnects $G$. Deleting all edges incident with a minimum-degree vertex proves $\lambda(G)\leq\delta(G)$. For a minimum edge cut separating vertex sets $A,B$, its endpoints on either suitable side give a vertex separator of size at most the number of cut edges; the complete-graph convention gives the same conclusion. Hence
$$
\delta(G)\geq\lambda(G)\geq\kappa(G).
$$
To realize prescribed $d\geq\ell\geq k\geq1$, take two disjoint copies $A,B$ of $K_{d+1}$. Add a bipartite set of exactly $\ell$ cross edges whose maximum matching has size $k$: include $a_ib_i$ for $1\leq i\leq k$, and, when $\ell>k$, add $a_1b_j$ for $k<j\leq\ell$. This is simple because $\ell\leq d$, has a matching of size $k$, and all its edges are covered by $\{a_1,\ldots,a_k\}$.
Vertices untouched by cross edges have degree $d$, so $\delta(G)=d$. The cross edges form an edge cut of size $\ell$; every cut that splits either clique uses at least $d\geq\ell$ clique edges, hence $\lambda(G)=\ell$. By König's theorem, the cross graph has a vertex cover of size $k$, whose deletion separates the two surviving clique pieces. Deleting fewer than $k$ vertices leaves each clique connected and at least one cross edge, so $\kappa(G)=k$.
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