7. Given a regular quadrilateral pyramid $P-ABCD$ with base edge length $AB=2$, height $PO=3$. $O'$ is a point on the line segment $PO$, and a plane parallel to the base of the regular quadrilateral pyramid $P-ABCD$ is drawn through $O'$, intersecting the edges $PA, PB, PC, PD$ at points $A', B', C', D'$ respectively. Then the maximum volume of the quadrilateral pyramid $O-A'B'C'D'$ is $\qquad$ .
7. $\frac{16}{27}$.

Let point $A^{\prime}$ be the projection of point $Q$ on line segment $A O$. Suppose the height of the pyramid $O-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is $x(0<x<3)$.
Given $A B=2 \Rightarrow A C=2 \sqrt{2} \Rightarrow A O=\sqrt{2}$.
By $\triangle A A^{\prime} Q \backsim \triangle A P O$
$$
\begin{array}{l}
\Rightarrow \frac{x}{A Q}=\frac{3}{\sqrt{2}} \Rightarrow A Q=\frac{\sqrt{2}}{3} x \\
\Rightarrow A^{\prime} O^{\prime}=\sqrt{2}-\frac{\sqrt{2}}{3} x \\
\Rightarrow A^{\prime} B^{\prime}=2-\frac{2}{3} x .
\end{array}
$$

Thus, the area of the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is
$$
S=\left(2-\frac{2}{3} x\right)^{2}=4-\frac{8}{3} x+\frac{4}{9} x^{2} \text {, }
$$

The volume of the pyramid $O-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is
$$
\begin{array}{l}
V=\left(4-\frac{8}{3} x+\frac{4}{9} x^{2}\right) x \times \frac{1}{3} \\
=\frac{4}{27} x^{3}-\frac{8}{9} x^{2}+\frac{4}{3} x .
\end{array}
$$

Using a TI calculator, it can be calculated that the maximum value is $\frac{16}{27}$ when $x=1$, and $V=0$ when $x=3$.
Therefore, the maximum value of $V$ is $\frac{16}{27}$.