Two tuning forks when sounded together produced 4 beats/sec. The frequency of one fork is 256. The number of beats heard increases when the fork of frequency 256 is loaded with wax. The frequency of the other fork is 

Suppose two tuning forks are named A and B with frequencies ${\mathrm{n}}_{\mathrm{A}}=256\mathrm{Hz}$ (known), nB = ? (unknown), and beat frequency x = 4 bps.

Frequency of unknown tuning fork may be

${\mathrm{n}}_{\mathrm{B}}=256+4=260\mathrm{Hz} \mathrm{or} =256-4=252\mathrm{Hz}$

It is given that on sounding waxed fork A (fork of frequency 256 Hz) and fork B, number of beats (beat frequency) increases. It means that with decrease in frequency of A, the difference in new frequency of A and the frequency of B has increased. This is possible only when the frequency of A while decreasing is moving away from the frequency of B.
This is possible only if n_B = 260 Hz. 
$\mathbf{Alternate}\mathbf{ }\mathbf{method}\mathbf{ }$: It is given  ${\mathrm{n}}_{\mathrm{A}}=256\mathrm{Hz},{\mathrm{n}}_{\mathrm{B}}=?$ and x = 4 bps
Also after loading A (i.e. n_A $\downarrow$), beat frequency (i.e. x) increases ($\uparrow$).
Apply these informations in two possibilities to known the frequency of unknown tuning fork.
         n_A $\downarrow$– n_B = x$\uparrow$        ... (i) 
         n_B  – n_A $\downarrow$= x$\uparrow$       ... (ii)
It is obvious that equation (i) is wrong (ii) is correct so 
n_B = n_A + x = 256 + 4 = 260 Hz.