We can use this law to calculate the approximate atomic weight of metal first.This is equal to
$\large\frac{6.4}{0.057}$$=112.28g/mol$
But this is not the value of X because $X$ is exact atomic weight.
From the data given it is possible to calculate the exact equivalent weight as
Mass of metal deposited =0.783g
Charge passed =$0.15\times 150\times 60$
$\Rightarrow 1350Coulombs$
Moles of electrons =$\large\frac{1350}{96500}$
$\Rightarrow 0.014$
$\therefore$ Equivalent weight=$\large\frac{0.783}{0.014}$
Now if we divide the approximate atomic weight,we will get the approximate valency of metal in salt.
$\therefore$ Approximate valency=$\large\frac{112.28}{55.93}$=2.007
Since valency has to be an integer,$n=2$
$\therefore $ Exact atomic weight $=55.93\times 2$
$\Rightarrow 111.86gm/mole$
Hence (d) is the correct answer.