The correct option is:
$20(3+\sqrt{14})$

Step-by-Step Explanation:

1. Identify the given values from the problem statement:
- Center of the circle: $C$
- Radius of the circle: $R=6\sqrt{2}\text{cm}$
- Angle: $\angle PQC={45}^{\circ }$
- The parallel chords $PQ$ and $SR$ are separated by one of the diameters (meaning they lie on opposite sides of the center $C$).

2. Determine the perpendicular distance of chord $PQ$ from the center $C$ ($d_1$):
In the triangle $△CPQ$, we have $CP=CQ=R$ (both are radii of the circle).
Since $△CPQ$ is an isosceles triangle:
$\angle CPQ=\angle PQC={45}^{\circ }$
This means the angle at the center is:
$\angle PCQ={180}^{\circ }-({45}^{\circ }+{45}^{\circ })={90}^{\circ }$
Therefore, $△CPQ$ is a right-angled isosceles triangle.
Let $d_1$ be the perpendicular distance from center $C$ to the chord $PQ$. Using trigonometry:
$d_1=R\sin \left({45}^{\circ }\right)=6\sqrt{2}\times \frac{1}{\sqrt{2}}=6\text{cm}$
The length of chord $PQ$ is:
$PQ=2\sqrt{R^2-d_1^2}=2\sqrt{72-36}=2\sqrt{36}=12\text{cm}$

3. Find the perpendicular distance of chord $SR$ from the center $C$ ($d_2$):
We are given that the ratio of the perpendicular distances of $PQ$ and $SR$ from $C$ is $3:2$.
$\frac{d_1}{d_2}=\frac{3}{2}⟹\frac{6}{d_2}=\frac{3}{2}⟹d_2=4\text{cm}$

4. Calculate the length of the chord $SR$:
Using the Pythagorean theorem for the chord $SR$:
$SR=2\sqrt{R^2-d_2^2}=2\sqrt{{\left(6\sqrt{2}\right)}^2-4^2}$
$SR=2\sqrt{72-16}=2\sqrt{56}=2\times 2\sqrt{14}=4\sqrt{14}\text{cm}$

5. Calculate the area of the quadrilateral $PQRS$:
Since $PQ$ and $SR$ are parallel, the quadrilateral $PQRS$ is a trapezoid.
Because the chords lie on opposite sides of the center, the height $h$ of the trapezoid is the sum of their perpendicular distances from $C$:
$h=d_1+d_2=6+4=10\text{cm}$
The area of the trapezoid is given by:
$\text{Area}=\frac{1}{2}\times (PQ+SR)\times h$
Substitute the values:
$\text{Area}=\frac{1}{2}\times (12+4\sqrt{14})\times 10$
$\text{Area}=5\times 4(3+\sqrt{14})$
$\text{Area}=20(3+\sqrt{14}){\text{cm}}^2$