1. Interpret the given statement

A block attached to a spring is pulled to $A = 10\text{ cm} = 0.10\text{ m}$ from equilibrium and released from rest.

Since it is released from rest at maximum displacement, this is the amplitude of SHM.

So, the total mechanical energy of the block-spring system is $E = \frac{1}{2}kA^2$

2. Use the given energy at $x=5\text{ cm}$

The question says: the energy of the block at $x = 5\text{ cm} = 0.05\text{ m}$ is $0.25\,$J.

In SHM, total mechanical energy remains constant at every position. Hence, $E = 0.25\text{ J}$

Thus, $\frac{1}{2}k(0.10)^2 = 0.25$

3. Solve for $k$

$\frac{1}{2}k(0.01) = 0.25$ $0.005k = 0.25$ $k = \frac{0.25}{0.005} = 50\,\text{N/m}$

4. Final answer

$\boxed{50}$

5. Compare with stored answer

Stored correct answer is $67$, but from the standard SHM energy relation, the spring constant is clearly $\boxed{50\,\text{N/m}}$

So I disagree with the stored answer. The likely issue is that the phrase "energy of the block" may have been misinterpreted in the stored key. If it means total mechanical energy of the block-spring system at that position, then the answer is definitely $50$. If it meant only kinetic energy at $x=5\,$cm, then: $K = E - U = \frac{1}{2}k(A^2-x^2)$ $0.25 = \frac{1}{2}k(0.10^2-0.05^2) = \frac{1}{2}k(0.0075)$ $k = \frac{0.25}{0.00375} \approx 66.7\,\text{N/m}$ which rounds to $67$.

Thus, the stored answer matches only if "energy of the block" is intended to mean kinetic energy of the block at $x=5$ cm, not total energy.