#### Answer

A,B,C,D,E,I,N,M,L,K,J,F,B,G,C,H,D,I,M,H,L,G,K,F,A is an Euler circuit.

#### Work Step by Step

We need to verify the number of odd vertices in the graph. Since all the vertices in this graph are even, the graph has zero odd vertices. Therefore, this connected graph has at least one Euler circuit.
To find an Euler circuit, we can start at any vertex.
Let's start at vertex A. According to Fleury's Algorithm, we should always choose an edge that is not a bridge, if possible. Since the edges AB and AF are not bridges, we can choose either of these edges as the next step in the path.
From vertex A, the path can travel to vertex B, then to vertex C, then to vertex D, then to vertex E, then to vertex I, then to vertex N, then to vertex M, then to vertex L, then to vertex K, then to vertex J, and then to vertex F.
At this point, we can see that the edge AF is a bridge, so according to Fluery's Algorithm, we must choose a different edge.
The path can then travel to vertex B, then to vertex G, then to vertex C, then to vertex H, and then to vertex D.
From there, the path must then travel to vertex I, then to vertex M, then to vertex H, then to vertex L, then to vertex G, then to vertex K, then to vertex F, and then finally back to vertex A, because these are the only available edges.
This path is A,B,C,D,E,I,N,M,L,K,J,F,B,G,C,H,D,I,M,H,L,G,K,F,A. This path travels through every edge of the graph exactly once, so it is an Euler path. Since it starts and ends at the same vertex, this path is an Euler circuit.
This is one Euler circuit but there are other Euler circuits in this graph also.