Answer: C

Video solution:

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Written solution:

Define $I$ to be the midpoint of $\overline{AD}$ and $G$ to be the midpoint of $\overline{EF}.$ Also define $H$ to be the intersection of $\overline{CF}$ and $\overline{BE}.$

The area of $\triangle BCE$ is $\dfrac12\cdot1\cdot4=2.$ By symmetry, we can see that $\triangle BCH$ and $\triangle EFH$ are similar. Since their bases are in a $1 : 3$ ratio, so are their altitudes. This means that $3IH = HG,$ which implies that $IH = 1.$

Therefore, $[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12.$ It follows that $[\triangle ECH]=2-\dfrac12=\dfrac32.$ Since the figure is symmetric, the total area of the bat wings is $2\cdot\dfrac32=3.$

Thus, C is the correct answer.