The elastic potential energy ($V$) stored in a spring is proportional to the square of its displacement ($x$) from its natural length, as defined by the formula $V(x) = \frac{1}{2}kx^2$, where $k$ is the spring constant .

1. Initial State: For a stretch of 2 cm, the energy is $U = \frac{1}{2}k(2)^2$.
2. Final State: For a stretch of 8 cm, the energy is $U' = \frac{1}{2}k(8)^2$.
3. Comparison: Dividing the two equations to find the relationship: $\frac{U'}{U} = \frac{\frac{1}{2}k(8)^2}{\frac{1}{2}k(2)^2} = \left(\frac{8}{2}\right)^2 = 4^2 = 16$

Thus, $U' = 16U$. Doubling the stretch quadruples the energy, and in this case, quadrupling the stretch (from 2 cm to 8 cm) increases the energy by a factor of $4^2 = 16$.