Let the radius of the cylinder be \( r \) decimeters. Then, the height of the cylinder is \( 2r \) decimeters. So, the volume of the cylinder is \[ \pi r^2 \cdot 2r = 2\pi r^3 \text{ cubic decimeters} \] If the height were \( 6r \) decimeters, then the volume would be \[ \pi r^2 \cdot 6r = 6\pi r^3 \text{ cubic decimeters} \] According to the condition: \[ 6\pi r^3 - 2\pi r^3 = 539 \Rightarrow 4\pi r^3 = 539 \Rightarrow 4 \times \frac{22}{7} \times r^3 = 539 \Rightarrow r^3 = \frac{539 \times 7}{4 \times 22} = \frac{49 \times 7}{4 \times 2} \Rightarrow r = \frac{7}{2} \] Therefore, the height of the cylinder is \[ \frac{7}{2} \times 2 = 7 \text{ decimeters} \quad \text{(Answer)} \]