$x^2-2\sqrt 3x+2=0$
Angle =$\large\frac{\pi}{3}$
We know that
$\cos c=\large\frac{a^2+b^2-c^2}{2ab}$
$\cos 60^{\large\circ}=\large\frac{a^2+b^2-c^2}{2ab}$
From the given equation we have
$a+b=2\sqrt 3$
$ab=2$
$\large\frac{1}{2}=\frac{a^2+b^2-c^2}{2ab}$
$\Rightarrow 2ab.\large\frac{1}{2}$$=a^2+b^2-c^2$
$ab=a^2+b^2-c^2$
$(a+b)^2-2ab-c^2=ab$
$(2\sqrt 3)^2-2\times 2-c^2=2$
$4\times 3-4-c^2=2$
$12-4-c^2=2$
$8-c^2=2$
$-c^2=2-8$
$c^2=6$
$c=\sqrt 6$
Perimeter =$a+b+c$
$\qquad\;\;\;\;\;=2\sqrt 3+\sqrt 6$
Hence (b) is the correct answer.