smc

All answers correctly grow as $m^2$, so we let $m=1$. The radius of the first circle is $\frac{1}{2}$, so its area is $\frac{\pi }{4}$. The diagonal of the second square is the diameter of the first circle, which is $1$. Therefore, the side length of the square is $\frac{\sqrt{2}}{2}$. Now we note that the picture is self-similar; if we erase the outer square, erase the outer circle, rotate the picture, and dilate the picture from the center in both the x- and y-directions by an equal scaling factor, we will get the original picture. Therefore, the side lengths of successive squares form a geometric sequence with common ratio $R$, as do the radii of the circles. On the other hand, the areas of the squares (and areas of the circles) form a geometric sequence with common ratio $R^2$. Since the first square has side $1$ and the second square has side $\frac{\sqrt{2}}{2}$, we know $R=\frac{\sqrt{2}}{2}$, and $R^2=\frac{1}{2}$. Since the area of the first circle is $\frac{\pi }{4}$, and the common ratio of areas is $\frac{1}{2}$, the sum of all the areas of the circles is the sum of the infinite geometric sequence, which is $\frac{A_1}{1-r}=\frac{\frac{\pi }{4}}{1-\frac{1}{2}}=\frac{\pi }{2}$. This is for side length $m=1$, and as noted before, there should be an $m^2$ factor in addition to this number to generalize it from the unit square. This gives answer $\boxed{\frac{\pi m^2}{2}}$.