Kuratowski's theorem states that a graph G is non-planar if and only if G has either $K_5$ or $K_{3,3}$ as a minor. Clearly this graph contains two $K_5$ minors, so our maximum non planar subgraph will involve at least removing one edge from each of the $K_5$ minors, call them $e_1, e_2$. We are then left with 24 - 2 = 22 edges. It remains to show that $G - e_1, e_2$ contains no $K_{3,3}$ minor. I am wondering if there is a way of accomplishing this just by using theorems, relations, and properties of the graphs, rather than visually attempting to uncover a $K_{3,3}$ minor as I find this relatively unintuitive and inefficient. For example, is there any way of using Euler's relation for planar graphs, V - E + F = 2 to solve this problem?
Here is a way to show that you cannot get a planar graph by removing two edges.
We know that a planar graph has at most $m = 3n-6$ edges, and it is attained only if the planar graph is triangulated.
If you remove an edge to the left $K_5$, it gives a triangulated graph, so no matter the embedding, its $5$ vertices will induce only $3$-sided faces.
No matter the edge you remove to the right $K_5$, its $5$ vertices will still be connected. So all these vertices must all belong to the same face of the graph induced by the $5$ other vertices, so in a triangle. However, the vertices of the right $K_5$ are adjacent to $4$ vertices in the left $K_5$, so the graph can't be planar.