1. The problem asks to find the 2nd term in the expansion of $\left(2x^3 - \frac{5}{x^2}\right)^4$ using the binomial theorem. 2. The binomial theorem states that for any integer $n$: $$\left(a + b\right)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. 3. Here, $a = 2x^3$, $b = -\frac{5}{x^2}$, and $n=4$. 4. The general term (the $(k+1)$-th term) is: $$T_{k+1} = \binom{4}{k} (2x^3)^{4-k} \left(-\frac{5}{x^2}\right)^k$$ 5. The 2nd term corresponds to $k=1$: $$T_2 = \binom{4}{1} (2x^3)^{3} \left(-\frac{5}{x^2}\right)^1$$ 6. Calculate the binomial coefficient: $$\binom{4}{1} = 4$$ 7. Calculate powers: $$(2x^3)^3 = 2^3 x^{3 \times 3} = 8x^9$$ $$\left(-\frac{5}{x^2}\right)^1 = -\frac{5}{x^2}$$ 8. Substitute back: $$T_2 = 4 \times 8x^9 \times \left(-\frac{5}{x^2}\right) = 4 \times 8 \times (-5) \times x^{9 - 2} = 4 \times 8 \times (-5) \times x^7$$ 9. Simplify the coefficients: $$4 \times 8 = 32$$ $$32 \times (-5) = -160$$ 10. Final term: $$T_2 = -160 x^7$$ Therefore, the 2nd term in the expansion is **$-160 x^7$**. The correct answer is option B.