${\mathrm{FeSO}}_4 + 2 \mathrm{NaOH}\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_2+{\mathrm{Na}}_2{\mathrm{SO}}_4. \mathrm{Equivalent} \mathrm{weight} \mathrm{of} {\mathrm{FeSO}}_4\mathrm{in} \mathrm{the} \mathrm{given} \mathrm{reaction} \mathrm{is}$$\left(\mathrm{M}=\mathrm{molar} \mathrm{mass} \mathrm{of} {\mathrm{FeSO}}_4\right)$

The given reaction is not a redox reaction.

 ${\mathrm{FeSO}}_4$  is acting like a salt. Valency of Fe in the salt  = 2 $\left({\mathrm{Fe}}^{+2}\right)$.

Valency of a salt = net charge of cation or net charge of anion

Equivalent Weight =$\frac{M}{2}$