Let the side of the equilateral triangle be $x$ m.

Then its perimeter is $3x$.

Since the total wire length is $22$ m, the remaining wire for the square is $22-3x.$

Express the side of the square in terms of $x$.

If the side of the square is $s$, then $4s=22-3x \,\Rightarrow\, s=\frac{22-3x}{4}.$

Write the total area as a function of $x$.

• Area of the equilateral triangle: $A_1=\frac{\sqrt{3}}{4}x^2$
• Area of the square: $A_2=s^2=\left(\frac{22-3x}{4}\right)^2$

So total area is $A(x)=\frac{\sqrt{3}}{4}x^2+\left(\frac{22-3x}{4}\right)^2.$

Differentiate to minimize.

$A(x)=\frac{\sqrt{3}}{4}x^2+\frac{(22-3x)^2}{16}$

Differentiate: $A'(x)=\frac{\sqrt{3}}{2}x+\frac{1}{16}\cdot 2(22-3x)(-3)$ $A'(x)=\frac{\sqrt{3}}{2}x-\frac{3}{8}(22-3x).$

For minimum, set $A'(x)=0$: $\frac{\sqrt{3}}{2}x-\frac{3}{8}(22-3x)=0.$

Multiply by $8$: $4\sqrt{3}x-3(22-3x)=0$ $4\sqrt{3}x-66+9x=0$ $x(9+4\sqrt{3})=66$ $x=\frac{66}{9+4\sqrt{3}}.$

Check that this gives a minimum.

$A''(x)=\frac{\sqrt{3}}{2}+\frac{9}{8}>0,$ so the area is indeed minimum.

Conclusion

The side of the equilateral triangle should be $\boxed{\frac{66}{9+4\sqrt{3}}}.$

Hence, the correct option is B.