Potassium ferrocyanide, ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$, has a central iron ion that is coordinated with six cyanide ligands in an octahedral geometry. The oxidation state of the iron ion in potassium ferrocyanide is $+2$, hence has ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridised state. On applying $\mathrm{C}.\mathrm{F}.\mathrm{T}$. it shows that all the $6$ electrons of $\mathrm{d}$ subshell are present in the form of three pairs in ${\mathrm{t}}_{2\mathrm{g}}$ orbitals. Hence answer is $3$.