$\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{3}{2}\Big)^6$ is equal to:

1. $\dfrac{8}{27}$

2. $\dfrac{27}{8}$

3. $\dfrac{4}{9}$

4. $\dfrac{9}{4}$

$\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{3}{2}\Big)^6\\[1em] = \Big(\dfrac{2 \times 2 \times 2}{3 \times 3 \times 3}\Big) \times \Big(\dfrac{3 \times 3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2 \times 2}\Big)\\[1em] = \Big(\dfrac{8}{27}\Big) \times \Big(\dfrac{729}{64}\Big)\\[1em] = \Big(\dfrac{8 \times 729}{27 \times 64}\Big)\\[1em] = \dfrac{5832}{1728} \\[1em] = \dfrac{27}{8}$

Hence, option 2 is the correct option.