1:2

As we can see, we can divide the side in the proportion 1:1:1, and it will look like this. There are 4 triangles, that are the half of the rectangle where they are. And there's also a rectangle in the middle, whose area is $\frac{1}{3}l \times \frac{1}{3}l = \frac{1}{9} A_{ABCD}$ And the triangles are the other $\frac{8}{9}$ , so it's $\frac{1}{2} \times \frac{8}{9} + \frac {1}{9} = \frac{5}{9}$

Let the height and breadth of rectangle $ABCD$ be $3a$ and $3b$ respectively. Then the area of quadrilateral $PQSR$ is:

$\begin{aligned} [PQRS] & = [ABCD] - [APS] - [BPQ] - [CQR] - [DRS] \\ & = (3a)(3b) - \frac 12(2a)b - \frac 12a(2b) - \frac 12(2a)b - \frac 12a(2b) \\ & = 9ab - ab - ab -ab - ab = 5ab \\ \implies \frac {[PQRS]}{[ABCD]} & = \boxed{\frac 59}\end{aligned}$