1. **State the problem:** We are given that the value of $k^{-5}$ is twice the value of $4k^{-2}$. We need to find the value of $k$. 2. **Write the equation:** According to the problem, $$k^{-5} = 2 \times 4k^{-2}$$ which simplifies to $$k^{-5} = 8k^{-2}$$ 3. **Rewrite the equation:** Divide both sides by $k^{-2}$ (assuming $k \neq 0$): $$\frac{k^{-5}}{k^{-2}} = 8$$ Using the rule $\frac{a^m}{a^n} = a^{m-n}$, we get $$k^{-5 - (-2)} = k^{-3} = 8$$ 4. **Solve for $k$:** $$k^{-3} = 8$$ Rewrite as $$\frac{1}{k^3} = 8$$ Taking reciprocal, $$k^3 = \frac{1}{8}$$ 5. **Find $k$:** $$k = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$$ 6. **Answer:** The value of $k$ is $\frac{1}{2}$, which corresponds to option B. **Final answer:** $k = \frac{1}{2}$