jmc

Since $△FGH$ is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram. All triangles are right-isosceles triangles. $CD=\frac{CG}{2}=3,DE=\frac{CD}{2}=\frac{3}{2},EF=\frac{DE}{2}=\frac{3}{4}$ $CB=CD=3,DK=DE=\frac{3}{2},EL=EF=\frac{3}{4}$ $[CBD]=\frac{1}{2}{3}^2=\frac{9}{2}$ $[DKE]=\frac{1}{2}(\frac{3}{2}{)}^2=\frac{9}{8}$ $[ELF]=\frac{1}{2}(\frac{3}{4}{)}^2=\frac{9}{32}$ The sum of the shaded regions is $\frac{9}{2}+\frac{9}{8}+\frac{9}{32}=\frac{189}{32}\approx 5.9$ $5.9$ is an underestimate, as some portion (but not all) of $△FGH$ will be shaded in future iterations. If you shade all of $△FGH$, this will add an additional $\frac{9}{32}$ to the area, giving $\frac{198}{32}\approx 6.2$, which is an overestimate. Thus, $6\to \boxed{6}$ is the only answer that is both over the underestimate and under the overestimate.