It is clear from the question that the centers of all three figures coincide. 

Let x be the length side of the equilateral triangle, hence the area (A) is ($$\sqrt{\ 3}$$/4)$$x^2$$

From this, the radius of the inscribed circle is= ($$\sqrt{\ 3}$$/2)x * (1/3) = x/2$$\sqrt{\ 3}$$

So, the diameter is x/$$\sqrt{\ 3}$$

Now, the diameter of the circle coincides with the diagonal of the square.

Hence, assuming the sides of square to be y, we have y$$\sqrt{\ 2}$$= x/$$\sqrt{\ 3}$$

y= x/$$\sqrt{\ 6}$$

From this, the area of square (B) is $$y^2$$, which is $$x^2$$/6

Now, the ratio of A and B is = $$3\sqrt{\ 3}$$/2