Each side of the large square in the figure is trisected (divided into three equal parts).  The corners of an inscribed square are at these trisection points, as shown.  The ratio of the area of the inscribed square to the area of the large square is
[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((1,0)--(1,0.2)); draw((2,0)--(2,0.2)); draw((3,1)--(2.8,1)); draw((3,2)--(2.8,2)); draw((1,3)--(1,2.8)); draw((2,3)--(2,2.8)); draw((0,1)--(0.2,1)); draw((0,2)--(0.2,2)); draw((2,0)--(3,2)--(1,3)--(0,1)--cycle); [/asy]
$\text{(A)}\ \dfrac{\sqrt{3}}{3} \qquad \text{(B)}\ \dfrac{5}{9} \qquad \text{(C)}\ \dfrac{2}{3} \qquad \text{(D)}\ \dfrac{\sqrt{5}}{3} \qquad \text{(E)}\ \dfrac{7}{9}$
Since we are dealing with ratios, let the big square have sides of $3$ and thus an area of $3^2 = 9$.  Chosing a multiple of $3$ will avoid fractions in the rest of the answer.
To find the area of the inscribed square, subtract off the areas of the four triangles.  Each triangle has an area of $\frac{1}{2}\cdot 2\cdot 1 = 1$.
Thus, the area of the inscribed square is $9 - 4\cdot 1 = 5$, and the ratio of areas is $\frac{5}{9}$, giving option $\boxed{B}$.