I saw this task in a textbook. I would be sure that this is impossible (I could solve this if there could be more than two SPE or if it had be better for just one player), but they present it as if not.
My reasoning is that if there are have to be 2 SPE, one player must have a choice at some node (he can decide for any of two strategies because they yield the same outcome). But if it were to happen again, there would certainly be more than 2. So only one choice in the backward induction. But that gives both the players the same payoff.
Note that subgame perfection is more general than backward induction, as it applies also to games with imperfect information.
Consider a coordination game. For instance, $$\begin{bmatrix} 2,2 & 0,0 \\ 0,0 & 1,1 \\ \end{bmatrix}$$ Write it in extensive form and it will have two equilibria in pure strategies and one in mixed strategies. The two equilibria in pure strategies satisfy the condition that one is better than the other.
For a game with exactly two equilibria, use the special coordination game $$\begin{bmatrix} 1,1 & 0,0 \\ 0,0 & 0,0 \\ \end{bmatrix}$$