3 squares + 2 triangles

Let the side length of the small equilateral triangle and square be $1$ . Joint the top vertices of the two equilateral triangles with a line. Then we note that the side length of the large equilateral triangle is $1+2\sqrt 3$ . Since the area of of a same shaped figure is directly promotional to the square of linear dimension, the ratio of the area of the small equilateral triangle to the area of the large equilateral triangle is:

$t = \frac {\Delta_{\rm small}}{\Delta_{\rm large}} = \frac {1^2}{(1+2\sqrt 3)^2} = \dfrac 1{13+4\sqrt 3} \approx 0.050180139 \implies \lfloor 10^6t\rfloor = \boxed{50180}$

Let length of each side of $\triangle XYZ$ be equal to $a$ . Since the three squares share a common side with $\triangle XYZ$ , the length of each side the squares is also $a$ .

Notice that $ZE=ZF$ $\angle ZEB=ZFB=90^\circ$ $\implies \triangle BEZ\cong \triangle BFZ$ So we have $\angle ZBE=\angle ZBF=\frac{\angle B}{2}=30^\circ$ In $\triangle BFZ$ , $\tan 30^\circ=\frac{ZF}{BF}=\frac{a}{BF}$ $\implies BF=a\sqrt{3}$ The length of each side of $\triangle ABC$ is $AB=AG+GF+BF=a\sqrt{3}+a+a\sqrt{3}=a(2\sqrt{3}+1)$ Therefore, $t=\frac{[\triangle XYZ]}{[\triangle ABC]}=\frac{\frac{\sqrt{3}}{4}\cdot XY^2}{\frac{\sqrt{3}}{4}\cdot AB^2}=\left ( \frac{a}{a(2\sqrt{3}+1)} \right)^2=\frac{1}{(2\sqrt3+1)^2}$ $\implies t\approx 0.05018013 \implies \boxed{\lfloor 10^6\cdot t\rfloor = 50180}$