Proof

By hypothesis, $k\ge 3$ is a positive integer, and $G(V,E)$ is a simple, biconnected, planar graph with shortest cycle length $k.$
Let $\mathcal D$ be a planar drawing of $G.$
Since all cycles in $G$ have the length of at least $k,$ all faces in the planar drawing $\mathcal D$ have a face degree of at least $k.$
It follows from the handshaking lemma for planar graphs that $k|F|\le 2|E|.$
Using the Euler's formula for planar graphs, we have $|F|=|E|-|V|+2.$
It follows $k|E|-k|V|+2k\le 2|E|,$ which is equivalent to $|E|\le \frac{k}{k-2}(|V|-2).$
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