1. **State the problem:** We have two similar solids (pyramids) with heights $h_1 = 4$ m and $h_2 = 2$ m. The volume of the smaller solid is $V_2 = 6$ m³. We need to find the volume $V_1$ of the larger solid. 2. **Formula and rules:** For similar solids, volumes scale as the cube of the ratio of their corresponding linear dimensions: $$\frac{V_1}{V_2} = \left(\frac{h_1}{h_2}\right)^3$$ 3. **Apply the formula:** Substitute the known values: $$\frac{V_1}{6} = \left(\frac{4}{2}\right)^3$$ 4. **Simplify the ratio inside the cube:** $$\frac{4}{2} = 2$$ So, $$\frac{V_1}{6} = 2^3 = 8$$ 5. **Solve for $V_1$:** $$V_1 = 6 \times 8 = 48$$ 6. **Final answer:** The volume of the larger solid is $\boxed{48}$ m³.