$\Big(-\dfrac{2a^2}{b^3}\Big)^4$ is equal to :

1. $-\dfrac{16a^8}{b^{12}}$

2. $\dfrac{16a^8}{b^{12}}$

3. $\dfrac{2a^8}{b^{12}}$

4. $-\dfrac{2a^8}{b^{12}}$

Simplifying the expression :

$\Rightarrow \Big(-\dfrac{2a^2}{b^3}\Big)^4 = \Big(-2 \times \dfrac{a^2}{b^3}\Big)^4 \\[1em] = (-2)^4 \times \Big(\dfrac{a^2}{b^3}\Big)^4 = 16 \times \dfrac{a^{2 \times 4}}{b^{3 \times 4}} \\[1em] = \dfrac{16a^8}{b^{12}}.$

Hence, Option 2 is the correct option.