That particular theorem is a "for all large enough $n$..." type result so I don't think it says much for small $n$. I'm not sure what would count as an *equivalent* formulation but it is interesting to ask why it has the form it does and what can be said about "if there are enough edges then their are cycles of a range of sizes."

With less than $n$ edges there need not be any cycles. For a connected graph with $n$ edges there is one cycle but it could be any size. With $n+1$ edges there are two cycles at least and a cycle using no more than about half the vertices (two bigger cycles would share vertices and create a smaller cycle.)

A complete bipartite graph has about $\frac{n^2}4$ edges but no odd cycles which is why the focus is on even length cycles.

Consider a graph with $n=2k+1$ made of $k$ triangles with a common vertex. There are no even length cycles and $\frac32 n -\frac12$ edges.

With at least $\frac32n$ edges (i.e. average degree at least $3$) There must be a sub-graph which is a *theta graph*, three internally disjoint paths sharing endpoints, and hence some even length cycle.

A hexagonal grid rolled up into a torus is regular of degree $3$ with cycles of lengths $6,10,12$ but not $4$ or $8.$ However this is not that optimal.

For any $d$ and $c$ there are graphs with minimal cycle length $c$ and all vertices of degree $d$. The number of vertices, $n$ grows quickly though. So what can be said for $n \approx 100?$

There is a result I've seen several place, such as here that there are graphs with $\frac{n}4(1+\sqrt{4n-3})$ edges must have $4$ cycles. That is average degree about $10.4$ when $n=100.$ I'm not sure how sharp that is.

There is a graph on $50$ vertices regular of degree $7$ with no $4$-cycles. One can put two of those together sharing a vertex to get $n=99$ and average degree just over $7.$

A projective plane of order $7$ has $57$ points and the same number of lines. This gives a graph with $n=114$, regular of degree $8$ and minimum cycle length $6.$ The corresponding affine plane gives a subgraph with $n=105$ and average degree $7 \frac{7}{15}$.