jmc

Let $P$ be the foot of the perpendicular from $F$ to the line containing $AB$. Since $\angle FAB=120^{\circ },$ then $\angle PAF=180^{\circ }-120^{\circ }=60^{\circ }$, and it follows that $△PAF$ is a $30-60-90$ triangle. As $AF=2$, it follows that $AP=1$ and $PF=\sqrt{3}$. Also, $AB=2$ and so $AX=3AB=6$. Thus, $PX=AP+AX=7$. In right triangle $FPX$, by the Pythagorean Theorem, it follows that $$F{X}^2=P{F}^2+P{X}^2=(\sqrt{3}{)}^2+(7{)}^2=52,$$and $FX=\sqrt{52}=\boxed{2\sqrt{13}}$.