The spin only magnetic moment for a complex can be calculated using formula

$\mathrm{\mu }=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}$

(a) ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$

The oxidation number $\mathrm{Fe}$ in ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$ is $+3$ .The electronic configuration of ${\mathrm{Fe}}^{3+}$ is $-\left(\mathrm{Ar}\right)3{\mathrm{d}}^{5}4{\mathrm{s}}^0$

As ${\mathrm{CN}}^{-}$ is a strong ligand pairing will occur.

${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$

$\therefore \mathrm{n}=1\Rightarrow \mathrm{\mu }=\sqrt{1\left(1+2\right)}=\sqrt{3}$

(b) ${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$

The oxidation number of $\mathrm{Fe}$ in

${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ is $+2$ .The electronic configuration of ${\mathrm{Fe}}^{2+}$ is $-\left[\mathrm{Ar}\right]3{\mathrm{d}}^{6}4{\mathrm{s}}^0$

As ${\mathrm{H}}_2\mathrm{O}$ is weak ligand, so pairing will not occur.

${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2-}$

$\mathrm{n}=4$

$\mathrm{\mu }=\sqrt{4\left(4+2\right)}=\sqrt{24}$

(c) ${\left[{\mathrm{MnF}}_6\right]}^{4-}$

The oxidation number of $\mathrm{Mn}$ is $+2$. It's

configuration is $-\left[\mathrm{Ar}\right]3{\mathrm{d}}^{5}4{\mathrm{s}}^0$

As ${\mathrm{F}}^{-}$ is a weak field ligand, so pairing will not occur.

${\left[{\mathrm{MnF}}_6\right]}^{4-}$

$\mathrm{n}=5$

$\mathrm{\mu }=\sqrt{5\left(5+2\right)}=\sqrt{35}$

(d) ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$

In ${\mathrm{NiCl}}_4^{2-}$, the oxidation number of $\mathrm{Ni}$ will be $+2.$ It's configuration is $-\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^0$

As ${\mathrm{Cl}}^{-}$ is a weak field ligand, so pairing will not occur.

${\left[{\mathrm{NiCl}}_4\right]}^{2-}$

$\mathrm{n}=2$

$\mathrm{\mu }=\sqrt{2\left(2+2\right)}=\sqrt{8}$

Thus, ${\left[{\mathrm{MnF}}_6\right]}^{4-}$ will have the highest spin only magnetic moment.