1. Set up the lengths

Let the wire be cut into two pieces of lengths $l_1$ and $l_2$. Since the total length is $20$ m, $l_1 + l_2 = 20.$

2. Area of square formed from length $l_1$

If a wire of length $l_1$ is bent into a square, then each side is $\frac{l_1}{4}.$ So its area is $A_1 = \left(\frac{l_1}{4}\right)^2 = \frac{l_1^2}{16}.$

3. Area of circle formed from length $l_2$

If a wire of length $l_2$ is bent into a circle, then $2\pi r = l_2 \implies r = \frac{l_2}{2\pi}.$ Hence area of the circle is $A_2 = \pi r^2 = \pi \left(\frac{l_2}{2\pi}\right)^2 = \frac{l_2^2}{4\pi}.$

4. Form the expression to be minimized

We need to minimize $2A_1 + 3A_2.$ Substitute $A_1$ and $A_2$: $2A_1 + 3A_2 = 2\cdot \frac{l_1^2}{16} + 3\cdot \frac{l_2^2}{4\pi}$ $= \frac{l_1^2}{8} + \frac{3l_2^2}{4\pi}.$ Using $l_2 = 20 - l_1$, $f(l_1) = \frac{l_1^2}{8} + \frac{3(20-l_1)^2}{4\pi}.$

5. Differentiate and equate to zero

$f'(l_1) = \frac{l_1}{4} + \frac{3}{4\pi}\cdot 2(20-l_1)(-1)$ $= \frac{l_1}{4} - \frac{3(20-l_1)}{2\pi}.$ For minimum, $f'(l_1)=0$ $\frac{l_1}{4} = \frac{3(20-l_1)}{2\pi}.$ Multiply by $4$: $l_1 = \frac{6(20-l_1)}{\pi}.$ Since $l_2 = 20-l_1$, $l_1 = \frac{6l_2}{\pi}.$ Therefore, $\pi l_1 = 6l_2.$ So, $(\pi l_1):l_2 = 6:1.$

6. Check that it is a minimum

$f''(l_1)=\frac{1}{4}+\frac{3}{2\pi}>0,$ so the critical point gives a minimum.

7. Final answer

$\boxed{(\pi l_1):l_2 = 6:1}$ So the correct option is A.