Step 1:
Equivalent weight of ferrous ammonium sulphate =392
$\therefore$ normality of salt solution =$\large\frac{3.92}{392}\times \frac{1000}{100}$
$\Rightarrow 0.1N$
$V_1N_1=V_2N_2$
$V_1N_1\Rightarrow $Mohr's salt
$V_2N_2\Rightarrow KmnO_4$
$20\times 0.1=18\times N_2$
$N_2=\large\frac{20\times 0.1}{18}$
$\Rightarrow \large\frac{1}{9}$$N$
Step 2:
Equivalent weight of $KMno_4=31.6$
$\therefore$ weight of $KMnO_4=31.6$
$\therefore$ Weight of $KMnO_4$ in litre=$31.6\times \large\frac{1}{9}$
$\Rightarrow 3.5$g
Hence (d) is the correct option.