Hexarea

The area of the equilateral triangle is $\frac{9\sqrt{3}}{4}cm^{2}$

The area of the three squares is $3\times9cm^{2}$ = $27cm^{2}$ .

The area of the two isosceles triangles is $\frac{3\times3\times\sin120º}{2}$ = $\frac{27\sqrt{3}}{4}cm^{2}$ .

Their total area is $27+9\sqrt{3}$ = $a+b\sqrt{c}$ .

a=27

b=9

c=3

And a - bc= $27-3\times3$ = $\boxed{0}$ .

Let $x$ be the area of each external triangle, $y$ be the area of each square, and $z$ be the area of the hexagon. Then

$3x=3\left(\dfrac{1}{2}(3)(3)\sin~120\right)=\dfrac{27}{2}\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{27}{4}\sqrt{3}$

$3y=3(3^2)=27$

$z=\dfrac{\sqrt{3}}{4}(3)(3)=\dfrac{9}{4}\sqrt{3}$

The area of the hexagon is therefore,

$3x+2y+z=\dfrac{27}{4}\sqrt{3}+27+\dfrac{9}{4}\sqrt{3}=27+9\sqrt{3}$

Finally, the desired answer is

$a-bc=27-(9)(3)=27-27=$ $\boxed{0}$