Given below are the half-cell reactions 

$\begin{array}{l}{\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mn}, {\mathrm{E}}^{\mathrm{o}}=-1.18\mathrm{V}\\ 2\left({\mathrm{Mn}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}\right), {\mathrm{E}}^{\mathrm{o}}=+1.51\mathrm{V}\end{array}$

The ${\mathrm{E}}^{\mathrm{o}}$ for $3{\mathrm{Mn}}^{2+}\to \mathrm{Mn}+2{\mathrm{Mn}}^{3+}$ will be 

$\begin{array}{l}{\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mn}, {\mathrm{E}}^{\mathrm{o}}=-1.18\mathrm{V}\\ 2\left({\mathrm{Mn}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}\right), {\mathrm{E}}^{\mathrm{o}}=+1.51\mathrm{V}\end{array}$

Subtracting Eq. (ii) from Eq. (i), we get 

$3{\mathrm{Mn}}^{2+}\to \mathrm{Mn}+2{\mathrm{Mn}}^{3+}$

${\mathrm{E}}^{\mathrm{o}}=-1.18-(+1.51)=-2.69\mathrm{V}$

Since, the value of ${\mathrm{E}}^{\mathrm{o}}$ is -ve, therefore the reaction is non-spontaneous. 

Alternate method 

$\begin{array}{l}{\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mn},{\mathrm{E}}^{\mathrm{o}}=-1.18\mathrm{V}...\left(\mathrm{i}\right)\\ {\mathrm{\Delta G}}^{\mathrm{o}}=-{\mathrm{nFE}}^{\mathrm{o}}\left[{\text{here, n = number of e}}^{-}\text{ inoved in the reaction}\right]\\ {\mathrm{\Delta G}}_1^{\mathrm{o}}=-2\mathrm{F}(-1.18)=2.36\mathrm{F}\\ 2{\mathrm{Mn}}^{3+}+2{\mathrm{e}}^{-}\to 2{\mathrm{Mn}}^{2+},{\mathrm{E}}^{\mathrm{o}}=+1.51\mathrm{V}...\left(\mathrm{ii}\right)\\ {\mathrm{\Delta G}}_2^{\mathrm{o}}=-2\mathrm{F}[-1.51]=-3.02\mathrm{F}\end{array}$

Subtracting Eq. (ii) from Eq. (i), we get 

$\begin{array}{l}3{\mathrm{Mn}}^{2+}\to \mathrm{Mn}+2{\mathrm{Mn}}^{3+},[\mathrm{n}=2]\\ {\mathrm{\Delta G}}_3^{\mathrm{o}}={\mathrm{\Delta G}}_1^{\mathrm{o}}-{\mathrm{\Delta G}}_2^{\mathrm{o}}=5.38\mathrm{F}\\ \Rightarrow -2{\mathrm{FE}}^{\mathrm{o}}=5.38\mathrm{F}\\ \Rightarrow {\mathrm{E}}^{\mathrm{o}}=-2.69\mathrm{V}\end{array}$