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} $
1. **Determine the inradius of the equilateral triangle:**
   - The area \( A \) of an equilateral triangle with side length \( s \) is given by:
     \[
     A = \frac{\sqrt{3}}{4} s^2
     \]
   - The semiperimeter \( s_p \) of the triangle is:
     \[
     s_p = \frac{3s}{2}
     \]
   - The area can also be expressed in terms of the inradius \( r \) and the semiperimeter:
     \[
     A = r \cdot s_p
     \]
   - Equating the two expressions for the area, we get:
     \[
     \frac{\sqrt{3}}{4} s^2 = r \cdot \frac{3s}{2}
     \]
   - Solving for \( r \):
     \[
     r = \frac{\frac{\sqrt{3}}{4} s^2}{\frac{3s}{2}} = \frac{\sqrt{3}}{4} s^2 \cdot \frac{2}{3s} = \frac{\sqrt{3}}{6} s
     \]

2. **Determine the diagonal of the inscribed square:**
   - The diameter of the circle (which is twice the radius) is:
     \[
     2r = 2 \cdot \frac{\sqrt{3}}{6} s = \frac{\sqrt{3}}{3} s
     \]
   - The diagonal of the inscribed square is equal to the diameter of the circle:
     \[
     \text{Diagonal of the square} = \frac{\sqrt{3}}{3} s
     \]

3. **Determine the side length of the inscribed square:**
   - Let \( a \) be the side length of the square. The relationship between the side length and the diagonal of a square is given by:
     \[
     a\sqrt{2} = \text{Diagonal}
     \]
   - Substituting the diagonal:
     \[
     a\sqrt{2} = \frac{\sqrt{3}}{3} s
     \]
   - Solving for \( a \):
     \[
     a = \frac{\frac{\sqrt{3}}{3} s}{\sqrt{2}} = \frac{\sqrt{3}}{3\sqrt{2}} s = \frac{\sqrt{3}}{3 \cdot \sqrt{2}} s = \frac{\sqrt{3}}{3 \cdot \sqrt{2}} s \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{3} \cdot \sqrt{2}}{3 \cdot 2} s = \frac{\sqrt{6}}{6} s
     \]

4. **Determine the area of the inscribed square:**
   - The area \( A_{\text{square}} \) of the square is:
     \[
     A_{\text{square}} = a^2 = \left( \frac{\sqrt{6}}{6} s \right)^2 = \frac{6}{36} s^2 = \frac{1}{6} s^2
     \]

Conclusion:
\[
\boxed{\frac{s^2}{6}}
\]