The n-factor for Ferrocyanide in the reaction given below: $\left[Fe{(CN{)}_6}^{4-}\right]⟶{Fe}^{3+}+{CO}_2+{{NO}_3}^{-1}$ is

The oxidation state of iron in reactant is $+2$ and in product is $+3$, net loss of electrons is $1$.

The oxidation state of carbon in reactant is $+2$ and in product is $+4$, net loss of electrons is $2$.

The oxidation state of nitrogen in reactant is $-3$ and in product is $+5$, net loss of electrons is $8$.

Overall reaction is oxidation.

There are $6$ carbons and nitrogen in the reactant, therefore the $n$-factor is

${\left[Fe(CN{)}_6\right]}^{4-}=\left[\right(1\times 1)+(6\times 2)+(6\times 8)=61$

Hence, the correct option is D.