A tuning fork with a frequency of 256 Hz generates 4 beats per...

Frequency of the tuning fork is $f_t = 256 \text{ Hz}$.

Since $4$ beats per second are heard, the possible frequencies of the wire are $f_w = 256 \pm 4$, which gives $260 \text{ Hz}$ or $252 \text{ Hz}$.

The fundamental frequency of a stretched wire is given by $f_w = \dfrac{v}{2L}$. As the length $L$ of the wire is shortened, its frequency $f_w$ increases.

It is given that upon shortening the wire, the beat frequency decreases. This is only possible if the initial frequency of the wire was lower than the tuning fork's frequency. Thus, the initial frequency of the wire is $f_{w1} = 252 \text{ Hz}$.

To produce zero beats, the new frequency of the wire must be equal to the frequency of the tuning fork, so $f_{w2} = 256 \text{ Hz}$.

Since the frequency is inversely proportional to the length ($f \propto \dfrac{1}{L}$), we have:

$f_{w1} L_1 = f_{w2} L_2$

$252 \times 25 = 256 \times L_2$

$L_2 = \dfrac{252 \times 25}{256} \approx 24.61 \text{ cm}$

The length by which the wire must be shortened is:

$\Delta L = L_1 - L_2 = 25 - 24.61 = 0.39 \text{ cm}$