jmc

Draw line $AD$, such that we create a larger triangle $△ABD$. $AC$ and $DE$ are medians of this triangle, and since all three medians of a triangle are concurrent, we can extend line $BF$ through $F$ to hit point $Q$ on line $AD$ such that $Q$ is the midpoint of $AD$.

The three medians of a triangle always divide the triangle into six smaller triangles of equal area. Knowing this, we have $[△AEF]=[△EFB]=[△FBC]=[△FCD]$. We see that $△ABC$ contains 3 of these smaller triangles. $BEFC$, our desired area, contains 2 of these smaller triangles. Hence $$[BEFC]=\frac{2}{3}[△ABC]=\frac{2}{3}\cdot \frac{2^2\sqrt{3}}{4}=\boxed{\frac{2\sqrt{3}}{3}}.$$