Problem 664

4. $\frac{\sqrt{7}}{2}$.

It is known that $S_{\triangle F E A}=3 S_{\triangle F B E}$, so $A E=3 B E$.
Similarly, we get $A F=4 C F$.
And $B E=2 C$, so we can set $C F=x$, then $A F=4 x, A C=$ $5 x, A E=6 x, A B=8 x$.
In $\triangle F E A$, using the cosine rule we get
$$E F^{2}=36 x^{2}+4^{2} x^{2}-2 \cdot 6 x \cdot 4 x \cdot \cos 60^{\circ}=28 x^{2} \text {. }$$

In $\triangle A B C$, using the cosine rule we get
$$\begin{aligned} B C^{2} & =64 x^{2}+25 x^{2}-2 \cdot 8 x \cdot 5 x \cdot \cos 60^{\circ} \\ & =49 x^{2} . \end{aligned}$$
$$\text { Then } \frac{B C}{E F}=\frac{7 x}{2 \sqrt{7} x}=\frac{\sqrt{7}}{2} \text {. }$$