We need to compare $\(\dfrac{2+\sqrt{3}}{(2-\sqrt{3})^3}\)$ with $\(\frac{10^6}{3^6}\)$
Column A has

$$
\( \frac{2+\sqrt{3}}{(2-\sqrt{3})^3}=\frac{2+\sqrt{3}}{(2-\sqrt{3})^3} \times \frac{(2+\sqrt{3})^3}{(2+\sqrt{3})^3}=\frac{(2+\sqrt{3})^2 \times(2+\sqrt{3})^2}{\left(2^2-\sqrt{3}^2\right)^3}=\frac{(7+4\sqrt{3})^2}{1^3}=(7+4 \sqrt{3})^2 \\
& =(7+4 \times 1.73)^2=(7+6.93)^2=(13.93)^2 \\
& \text { Column B } \frac{10^6}{3^6}=(3.33)^6=\left(3.33^3\right)^2=(36.93)^2
\)
$$


Clearly column B has greater quantity, so the answer is (B).