(a) $[\sqrt{3}(\sqrt{3}+1)]^{2}=3(3+1+2 \sqrt{3})$
$=12+6 \sqrt{3}$
(b) $[-\sqrt{3}(\sqrt{3}-1)]^{2}=3(3+1-2 \sqrt{3})$
$=12-6 \sqrt{3}$
(c) $[\sqrt{3}(-\sqrt{3}+1)]^{2}=3(3+1-2 \sqrt{3})$
$=12-6 \sqrt{3}$
Hence, none of the square values of options are equal to the given value.
Alternative
Let $\sqrt{6+4 \sqrt{3}}=\sqrt{x}+\sqrt{y}$
$\Rightarrow 6+4 \sqrt{3}=x+y+2 \sqrt{x y}$
$\Rightarrow x+y=6, \sqrt{x y}=2 \sqrt{3}$
Now, $(x-y)^{2}=(x+y)^{2}-4 x y=36-4(4 \times 3)$
$=-12 < 0$
It is not possible.
Hence, square root is not possible.