Adding Standard Reduction Potentials
It's somewhat unclear what you mean by "using $\Delta G$", but here's the full working. The information you are given is:


  $$\begin{align}
\ce{MnO4- + e-} &amp;\rightarrow \ce{MnO4^2-} &amp; E &amp;= +0.564~\mathrm{V} \tag{1}  \\
\ce{MnO4^2- + 2e- + 4H+} &amp;\rightarrow \ce{MnO2 + 2H2O} &amp; E &amp;= +2.261~\mathrm{V} \tag{2}
\end{align}$$


Converting these to Gibbs free energy changes is fundamentally the correct step. We have $\Delta G = -nFE$. (Standard state symbols omitted for clarity, they are implied throughout.) For half-equation (1), $n = 1$, and for half-equation (2), $n = 2$. So:

$$\begin{align}
\Delta G_1 &amp;= -(1)(96485~\mathrm{C~mol^{-1}})(0.564~\mathrm{V}) = -54417.7~\mathrm{J~mol^{-1}} \\
\Delta G_2 &amp;= -(2)(96485~\mathrm{C~mol^{-1}})(2.261~\mathrm{V}) = -436306.7~\mathrm{J~mol^{-1}} \\
\end{align}$$

You need to now sum up these two equations, which also means summing up the changes in Gibbs free energy. Let's define a new reaction $(3) = (1) + (2)$:


  $$\begin{align}
\ce{MnO4- + 3e- + 4H+} &amp;\rightarrow \ce{MnO2 + 2H2O} &amp; E &amp;= E_3 \tag{3}
\end{align}$$


As you can see, all I have done is to add the two half-equations together and cancel out any terms that appear on both sides. We have $n = 3$ for this half-reaction.

$$\begin{align}
\Delta G_3 &amp;= \Delta G_1 + \Delta G_2 \\
&amp;= -490724.4~\mathrm{J~mol^{-1}} \\
E_3 &amp;= -\frac{\Delta G_3}{nF} \\
&amp;= 1.695~\mathrm{V}
\end{align}$$

Last thing, please always quote your units in your answer! You cannot have an electrode potential that is unitless.