Ireland

Let $x=|BD|=|DC|$. Because $△ABC$ is equilateral, considering the areas of $△BDF$ and $△DCE$ gives $$54\sqrt{3}=\frac{1}{2}x\cdot |BF|\cdot \sin (60^{\circ })=\frac{1}{4}\sqrt{3}\cdot x\cdot |BF|\Rightarrow x\cdot |BF|=216$$ $$24\sqrt{3}=\frac{1}{2}x\cdot |CE|\cdot \sin (60^{\circ })=\frac{1}{4}\sqrt{3}\cdot x\cdot |CE|\Rightarrow x\cdot |CE|=96.$$

Because $|AF|=2x-|BF|$ and $|AE|=2x-|CE|$ we obtain for the area of $△AFE$ the equation $24\sqrt{3}=\frac{1}{2}(2x-|BF|)(2x-|CE|)\sin (60^{\circ })$ from which we get $96x^2=(2x^2-x\cdot |BF|)(2x^2-x\cdot |CE|)=(2x^2-216)(2x^2-96)$. This easily translates into the equation $0=x^4-108x^2+5184=(x^2-144)(x^2-36)$. Because $0\le |AF|=2x-\frac{216}{x}$, we need to have $2x^2\ge 216$, hence $x^2\ne 36$. As $x>0$ we see that the only solution is $x=12$. Therefore, $|AB|=|BC|=|CA|=24$.