1. Initial kinetic energy

The block initially has kinetic energy $K_i = E.$

2. When speed is halved

If the initial speed is $u$, then initial kinetic energy is $E = \frac{1}{2}mu^2.$

When the speed becomes $\dfrac{u}{2}$, the kinetic energy becomes $K_f = \frac{1}{2}m\left(\frac{u}{2}\right)^2 = \frac{1}{2}m\cdot \frac{u^2}{4} = \frac{1}{4}\left(\frac{1}{2}mu^2\right)=\frac{E}{4}.$

3. Loss in kinetic energy = gain in spring potential energy

Compression of spring is given as $x = 25\text{ cm} = 0.25\text{ m}.$

The decrease in kinetic energy is $\Delta K = E - \frac{E}{4} = \frac{3E}{4}.$

This is stored in the spring: $\frac{1}{2}kx^2 = \frac{3E}{4}.$

Substitute $x=0.25$ m: $\frac{1}{2}k(0.25)^2 = \frac{3E}{4}.$

Since $(0.25)^2 = \frac{1}{16},$ we get $\frac{1}{2}k\cdot \frac{1}{16} = \frac{k}{32} = \frac{3E}{4}.$

Hence, $k = 32\cdot \frac{3E}{4} = 24E.$

4. Compare with given form

Given, $k = nE\; \text{N m}^{-1}.$ So, $n=24.$

5. Comparison with stored answer

Stored correct answer = $24$.

Our derived answer also gives $n=24.$ So the stored answer is correct.