The answer is equal to \(148\).
Let \(G\) be the solution graph. Since \(G\) has no even cycle, no two cycles of \(G\) share any edge. In other words, the cycles of \(G\) are edge-disjoint. To show a bound on the number of edges of \(G\), we construct a graph \(H\) in the following way:

• For each vertex of \(G\) there is a corresponding vertex in \(H\).

• For each cycle of \(G\), there is a vertex in \(H\).

• For each edge \((u,v)\) of \(G\) that does not belong to any cycle, we put an edge between the corresponding vertices of \(u\) and \(v\) in \(H\).

• For each edge \((u,v)\) of \(G\) that belongs to a cycle \(c\), we put an edge between the corresponding vertex of \(c\) in \(H\) and both \(u\) and \(v\). We remove duplicate edges so \(H\) is also simple.

It follows from the construction of \(H\) and the fact that cycles of \(G\) are edge disjoint that (i) \(H\) is a tree and (ii) the number of edges of \(H\) is equal to the number of edges of \(G\). This implies that the number of edges of \(G\) is equal to \(|V(H)-1|\) where \(|V(H)|\) is the number of vertices of \(H\). Since every cycle of \(G\) has a length of at least \(3\), the degrees of all vertices of \(H\) that correspond to cycles of \(G\) are at least \(3\). Moreover, no two such vertices of \(H\) are adjacent and thus their count is bounded by \(\lfloor (|V(H)|-1)/3 \rfloor\). \(|V(G)| =100\) implies that \(|V(H)| \leq 149\) and since \(H\) is a tree then it has at most \(148\) edges.

It is easy to see that by adding 49 edges to 98 disjoint pairs of leaves of a star with 100 vertices, we obtain a simple graph with 100 vertices and 148 edges that has no even cycle.