The answer is $\lfloor \frac{3}{2}(n-1)\rfloor$.  First note that if $G$ is $2$-connected and even-cycle-free, then $G$ must just be an odd cycle.  To see this, consider an [ear-decomposition](https://en.wikipedia.org/wiki/Ear_decomposition) of $G$.  If $G$ is not just a cycle, then $G$ contains a cycle and an ear. However, by parity considerations, a cycle and an ear always contains an even cycle.  

Now let $G$ be an arbitrary even-cycle-free graph and consider its [block-cut-tree](https://en.wikipedia.org/wiki/Biconnected_component) $T$.  By the previous remark, each block of $G$ is an odd cycle or just an edge.  To maximize the number of edges, each block of $G$ should be a triangle.  Thus, the maximum number of edges is attained by a 'tree of $k$ triangles.' This graph has $3k$ edges and $2k+1$ vertices.