#### Answer

B,C,F,I,H,G,D,A,B,E,F,E,H,E,D,E,B is an Euler circuit.

#### Work Step by Step

Let's start at vertex B. According to Fleury's Algorithm, we should always choose an edge that is not a bridge, if possible. Since the edges AB, BC, BE, and BE are not bridges, we can choose any of these edges as the next step in the path.
From vertex B, one option is travel all the way around the outside of the graph. The path can travel to vertex C, then to vertex F, then to vertex I, then to vertex H, then to vertex G, then to vertex D, then to vertex A, and then back to vertex B.
At this step, the only option is to travel to vertex E. After this step, we can see that edge BE is a bridge, so according to Fluery's Algorithm, we must choose a different edge.
The path can travel to vertex F, and then back to vertex E. Then the path can travel to vertex H, and then back to vertex E.
At this step, we can see that the edge BE is a bridge, so according to Fluery's Algorithm, we must choose a different edge.
The path must travel to vertex D. After this step, the path must then travel back to vertex E, and then finally back to vertex B, because these are the only available edges.
This path is B,C,F,I,H,G,D,A,B,E,F,E,H,E,D,E,B. This path travels through every edge of the modified 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 modified graph also.