The correct order of spin only magnetic moments for the following complex ions is

${\left[{\mathrm{MnBr}}_4\right]}^{2-}$ is a tetrahedral complex. The electronic configuration ${\mathrm{Mn}}^{2+}$is${\mathrm{e}}^2{\mathrm{t}}_2^3$ and it has $5$ unpaired electrons.

$\therefore \mathrm{\mu }=\sqrt{5(5+2)}=\sqrt{35}\text{ B. }\mathrm{M}.$

${\left[{\mathrm{CoF}}_6\right]}^{3-}$ is an octahedral complex and it is high spin complex. The electronic configuration of ${\mathrm{Co}}^{3+}$ is  $t_{2g}^4{e_g}^2$, it has $4$ unpaired electrons

$\therefore \mathrm{\mu }=\sqrt{4(4+2)}=\sqrt{24}\text{ B.M. }$

${\left[\mathrm{Mn}(\mathrm{CN}{)}_6\right]}^{3-}$ is an octahedral complex and it is low spin complex. The electronic configuration ${\mathrm{Mn}}^{3+}$ is $t_{2g}^4{e}_g^0$ it has two unpaired electrons

$\therefore \mu =\sqrt{2(2+2)}=\sqrt{8}\text{ B.M }$

$\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6{]}^{3-}$ is an octahedral complex and it is low spin complex. The electronic configuration of ${\mathrm{Fe}}^{3+}$ is ${\mathrm{t}}_{2g}^5{\mathrm{eg}}^0$ and it has one unpaired electron.

$\therefore \mathrm{\mu }=\sqrt{1(1+2)}=2\sqrt{3}\text{ B.M }$

Hence the correct answer is option $3$