The equivalent weight of K₄[Fe(CN)₆] in the given reaction is:

${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]\stackrel{\left[0\right]}{⟶}{\mathrm{Fe}}^{3+}+{\mathrm{NO}}_3^{-}+{\mathrm{CO}}_2$

iron looses 1 e

six carbons looses 12 e

six nitrogens looses 48 e

Net change in oxidation number = + 60 + 1 = + 61 

$\text{ Eq. wt. }=\frac{M}{61}$