Explanation

(c) Suppose two tuning forks are named $A$ and $B$ with frequencies $n_{A} = 256\ Hz\ (known),\ n_{B} = ?$ (unknown), and beat frequency $x = 4\ bps.$
Frequency of unknown tuning fork may be
$n_{B} = 256 + 4 = 260\ Hz$ and $n_{B} = 256 - 4 = 252\ Hz$
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 $n_{B} = 260\ Hz.$
Alternate method : It is given $n_{A} = 256\ Hz,\ n_{B} = ?$ and $x = 4\ bps$
Also after loading $A\ \left( i.e.\ n_{A} \downarrow \right),$ beat frequency ($i.e.\ x)$ increases $( \uparrow$).
Apply these informations in two possibilities to known the frequency of unknown tuning fork.
$n_{A} \downarrow - n_{B} = x$ …(i)
$n_{B} - n_{A} \downarrow = x \uparrow$ …(ii)
It is obvious that equation (i) is wrong (ii) is correct so
$n_{B} = n_{A} + x = 256 + 4 = 260\ Hz$