$23 \cdot 11$ In a square $A B C D$ with side length 2, $E$ is the midpoint of $A B$. Now fold along $E C, E D$ so that $A E$ and $B E$ coincide, and points $A$ and $B$ coincide, named as point $O$. Then the volume of the tetrahedron $O-C D E$ is
(A) $\sqrt{3}$.
(B) $\frac{\sqrt{3}}{2}$.
(C) $\frac{\sqrt{3}}{3}$.
(D) $\frac{\sqrt{3}}{4}$.
[Solution] After folding, we get the tetrahedron as shown in the figure, with the base $O C D$ being an equilateral triangle with side length 2, so its area is $\sqrt{3}$.

Also, $O E \perp O C, O E \perp O D$, so $O E \perp$ plane $O C D$, meaning $E O$ is the height of the base $O C D$ of the tetrahedron, and $E O=1$.
Thus,
$$
\begin{aligned}
V_{O-C D E} & =\frac{1}{3} \cdot S_{\triangle O C D} \cdot h \\
& =\frac{1}{3} \cdot \sqrt{3} \cdot 1=\frac{\sqrt{3}}{3} .
\end{aligned}
$$

Therefore, the answer is $(C)$.