Let $$x \times y = x^2 + y^3$$ and $$(x \times 1) \times 1 = x \times (1 \times 1)$$. Then a value of $$2\sin^{-1}\left(\frac{x^4 + x^2 - 2}{x^4 + x^2 + 2}\right)$$ is

We are given the operation $$x \times y = x^2 + y^3$$, and the condition $$(x \times 1) \times 1 = x \times (1 \times 1)$$. We compute $$x \times 1 = x^2 + 1^3 = x^2 + 1$$ so the left side is $$(x^2 + 1) \times 1 = (x^2 + 1)^2 + 1^3 = (x^2 + 1)^2 + 1 = x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 + 2$$ while the right side is $$1 \times 1 = 1^2 + 1^3 = 2$$ and thus $$x \times 2 = x^2 + 2^3 = x^2 + 8$$. Setting these equal gives $$x^4 + 2x^2 + 2 = x^2 + 8$$ which simplifies to $$x^4 + x^2 - 6 = 0$$. Letting $$t = x^2$$ yields $$t^2 + t - 6 = 0$$, so $$(t + 3)(t - 2) = 0$$ and hence $$t = 2$$ (rejecting $$t = -3$$ since $$t = x^2 \ge 0$$), which means $$x^2 = 2$$ and $$x^4 = 4$$. Finally, the expression becomes $$\frac{x^4 + x^2 - 2}{x^4 + x^2 + 2} = \frac{4 + 2 - 2}{4 + 2 + 2} = \frac{4}{8} = \frac{1}{2}$$ and therefore $$2\sin^{-1}\left(\frac{1}{2}\right) = 2 \times \frac{\pi}{6} = \frac{\pi}{3}$$.

Therefore, the answer is Option B: $$\frac{\pi}{3}$$.