The radius of our circle is $\frac{\sqrt{3}}{6}s$. This means the square has side length $\frac{\sqrt{6}}{6}s$ which has area of $\frac{s^2}{6}=\fbox{B}$.
The side of an equilateral triangle is $s$.  A circle is inscribed in the triangle and a square is inscribed in the circle.  The area of the square is:
$\text{(A)}\ \frac{s^2}{24}\qquad\text{(B)}\ \frac{s^2}{6}\qquad\text{(C)}\ \frac{s^2\sqrt{2}}{6}\qquad\text{(D)}\ \frac{s^2\sqrt{3}}{6}\qquad\text{(E)}\ \frac{s^2}{3}$