As you said, $X$ is the Fermat point.
And also with having three equilateral triangles on the sides, we have an interesting fact that $AA'=BB'=CC'=L$. (Here's a good link where you may find the proof and some other interesting things)
Now consider $\Delta CXA'$, we could see $$CX=L-5$$$$XA'=L-3$$$$CA'=4$$
And the calculations will show us $L=7.6$, now we are interested in calculating $BX$ and $B'X$
By solving this system of equations and using
We can find out that $BX=2$, therefore $BX'=5.6$.
Now by calculating areas of $\Delta B'XC'$, $\Delta BXC$, $\Delta BXC'$ and $\Delta B'XC$ we may find the area of $BCB'C'$ ( forgive me if I skipped the calculation part in my solution! I hope the total vision is right)