Rhombus $ABCD$ is similar to rhombus $BFDE$. The area of rhombus $ABCD$ is $54$, and $\angle BAD=30^\circ$. What is the area of rhombus $BFDE$?

1. Let the side lengths of $ABCD$ and $BFDE$ be $s$ and $t$. In $ABCD$, $BD$ is the short diagonal, so $BD=2s\sin15^\circ$.
2. In the similar rhombus $BFDE$, $BD$ is the long diagonal, so $BD=2t\cos15^\circ$. Equating the diagonals gives $t/s=\tan15^\circ=2-\sqrt3$.
3. The area ratio is $(t/s)^2=(2-\sqrt3)^2=7-4\sqrt3$, so $[BFDE]=54(7-4\sqrt3)=\boxed{378-216\sqrt3}$.