 Solution

Every facial boundary walk has length at least five. Bridge excursions can repeat an edge, but deleting those excursions leaves a boundary containing a cycle in a graph; the finite girth assumption ensures the graph is not a tree. Thus the shortest-cycle condition still gives the lower bound five even when bridges occur. Summing boundary lengths counts each edge twice, so $5f\le 2m$. Using Euler formula for a connected planar graph,Hence