A tuning fork generates 4 beats per second when sounded with...

Let the frequency of the first tuning fork be $n_1$ and the second be $n_2 = 256 \text{ Hz}$.

The initial beat frequency is $4 \text{ Hz}$.

$n_1 = n_2 \pm 4 = 256 \pm 4$

$n_1 = 260 \text{ Hz}$ or $252 \text{ Hz}$

When the first tuning fork is loaded with wax, its frequency decreases.

If $n_1 = 260 \text{ Hz}$, a small decrease in frequency would initially decrease the beat frequency (for example, to $259 \text{ Hz}$, giving $3$ beats).

If $n_1 = 252 \text{ Hz}$, a small decrease in frequency would increase the beat frequency (for example, to $250 \text{ Hz}$, giving $6$ beats).

Since the beat frequency increases to $6 \text{ Hz}$, the original frequency of the first tuning fork must be $252 \text{ Hz}$.