The potential energy stored in a spring with spring constant $k$ that is stretched a distance $d$ from its equilibrium position is

$\begin{align*}U = \frac{1}{2}k d^2\end{align*}$
Let $k'$ denote the spring constant of the second spring. Then, from the problem statement,

$\begin{align*}2 \cdot \left(\frac{1}{2}k d^2\right) = \frac{1}{2} k' \left(\frac{d}{2}\right)^2\end{align*}$
Solving for $k'$ gives

$\begin{align*}k' = \boxed{8 k}\end{align*}$
Therefore, answer (D) is correct.

Solution to 2008 Problem 28