#### Answer

A,B,C,A,F,D,E,F,B,D,C,E,B is an Euler path.

#### Work Step by Step

We need to verify the number of odd vertices in the graph. Vertex A and vertex B are odd vertices. The other vertices are even. Since the connected graph has exactly two odd vertices, the graph has at least one Euler path.
If a graph has exactly two odd vertices, then any Euler path starts at one odd vertex and ends at the other odd 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 none of the edges AB, AC, or AF are bridges, we can choose any of these three edges as the next step in the path.
From vertex A, the path can travel to vertex B, then to vertex C, and then back to vertex A. From there, the path must travel to vertex F because it is the only available edge.
Since none of the edges BF, DF, or EF are bridges, we can choose any of these three edges as the next step in the path.
From vertex F, the path can travel to vertex D, then to vertex E, and then back to vertex F. From there, the path must travel to vertex B because it is the only available edge.
The path can then move to vertex D, and then the path must travel to vertex C, then to vertex E, and finally back to vertex B.
This path is A,B,C,A,F,D,E,F,B,D,C,E,B. This path travels through every edge of the graph exactly once, so it is an Euler path.
This is one Euler path but there are other Euler paths in this graph also.