Given $\triangle{ABC}$ with $\angle{B}=60^{\circ}$ and $\angle{C}=30^{\circ}$, let $P,Q,R$ be points on the sides $BA,AC,CB$ respectively such that $BPQR$ is an isosceles trapezium with $PQ \parallel BR$ and $BP=QR$.\\
Find the maximum possible value of $\frac{2[ABC]}{[BPQR]}$ where $[S]$ denotes the area of any polygon $S$.
1. **Determine the angles of $\triangle ABC$:**
   Given $\angle B = 60^\circ$ and $\angle C = 30^\circ$, we can find $\angle A$ using the fact that the sum of the angles in a triangle is $180^\circ$.
   \[
   \angle A = 180^\circ - \angle B - \angle C = 180^\circ - 60^\circ - 30^\circ = 90^\circ
   \]
   Therefore, $\triangle ABC$ is a right triangle with $\angle A = 90^\circ$.

2. **Identify the properties of the isosceles trapezium $BPQR$:**
   Since $BPQR$ is an isosceles trapezium with $PQ \parallel BR$ and $BP = QR$, we know that $PQ$ and $BR$ are the parallel sides, and $BP$ and $QR$ are the non-parallel sides of equal length.

3. **Express the areas of $\triangle ABC$ and $BPQR$:**
   Let $a = BC$, $b = CA$, and $c = AB$. Since $\triangle ABC$ is a right triangle with $\angle A = 90^\circ$, we can use the Pythagorean theorem:
   \[
   a^2 + b^2 = c^2
   \]
   The area of $\triangle ABC$ is:
   \[
   [ABC] = \frac{1}{2} \times b \times a
   \]

4. **Determine the area of $BPQR$:**
   Since $BPQR$ is an isosceles trapezium with $PQ \parallel BR$ and $BP = QR$, we can use the properties of trapeziums to find its area. Let $h$ be the height of the trapezium from $PQ$ to $BR$. The area of $BPQR$ is:
   \[
   [BPQR] = \frac{1}{2} \times (PQ + BR) \times h
   \]

5. **Relate the areas of $\triangle ABC$ and $BPQR$:**
   To find the maximum possible value of $\frac{2[ABC]}{[BPQR]}$, we need to express $[BPQR]$ in terms of $[ABC]$. Since $BPQR$ is inscribed in $\triangle ABC$, the maximum area of $BPQR$ occurs when $BPQR$ occupies the largest possible area within $\triangle ABC$.

6. **Calculate the maximum possible value:**
   The maximum area of $BPQR$ occurs when $BPQR$ is similar to $\triangle ABC$ and scaled down by a factor. Since $BP = QR$, the maximum area of $BPQR$ is half the area of $\triangle ABC$:
   \[
   [BPQR] = \frac{1}{2} [ABC]
   \]
   Therefore, the maximum possible value of $\frac{2[ABC]}{[BPQR]}$ is:
   \[
   \frac{2[ABC]}{\frac{1}{2}[ABC]} = \frac{2[ABC]}{[ABC]/2} = 4
   \]

The final answer is $\boxed{4}$.