Answer: $\frac{2}{9}$

Solution: Let the midpoints of sides $AB, BC, CD,$ and $DA$ of quadrilateral $ABCD$ be denoted by $X, Y, Z,$ and $T$ respectively. By the property of medians, points $P, Q, R, S$ divide segments $OX, OY, OZ, OT$ in the ratio 2:1, counting from $O$. Therefore, the area of quadrilateral $PQRS$ is 
$\left(\frac{2}{3}\right)^{2}=\frac{4}{9}$ of the area of quadrilateral $XYZT$. We will show that the area of the latter quadrilateral is half the area of $ABCD$. Note that segment $XY$ is the midline of triangle $ABC$, so the area of triangle $XBY$ is one-fourth the area of triangle $ABC$. Similarly, the area of triangle $ZDT$ is one-fourth the area of triangle $CDA$, and the sum of the areas of $XBY$ and $ZDT$ is one-fourth the area of $ABCD$. The same can be shown for the sum of the areas of $YCZ$ and $TAX$, which is also one-fourth the area of $ABCD$. Finally, the area of $XYZT$ is the difference between the area of $ABCD$ and the areas of triangles $XBY, ZDT, YCZ,$ and $TAX$, which is half the area of $ABCD$.

Ultimately, the area of quadrilateral $PQRS$ is $\quad \frac{1}{2}\left(\frac{2}{3}\right)^{2}=\frac{2}{9} \quad$ of the area of quadrilateral $ABCD$.

Grading criteria: Proved that the area of quadrilateral $PQRS$ is $\left(\frac{2}{3}\right)^{2}=\frac{4}{9}$ of the area of quadrilateral $XYZT$: 4 points. Proved that the area of quadrilateral $XYZT$ is half the area of $ABCD$: 3 points.
11.4. Let $\mathrm{O}$ be the point of intersection of the diagonals of a convex quadrilateral $A B C D$, and let $P, Q, R$, $S$ be the points of intersection of the medians of triangles $A O B, B O C, C O D$ and $D O A$ respectively. Find the ratio of the areas of quadrilaterals $P Q R S$ and $A B C D$.