As correct as eflaschuks approach is, I found it rather confusing, because much text and few mathematics. So I am adding them here.
First of all, let us write down the Nernst Equation for an arbitrary reaction $\ce{Ox + z\cdot e- <=> Red}$:
\begin{aligned}
E=E^\ominus(\ce{Red/Ox}) - z^{-1}\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Red}}{\ce{Ox}}
\end{aligned}
For our given system that boils down to the following reaction and their potentials:
\begin{aligned}
\ce{Fe^{2+} + $2\cdot$ e- &-> Fe}& E_1 &= E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}& \text{(1)}\\
\ce{Fe^{3+} + $1\cdot$ e- &-> Fe^{2+}}& E_2 &= E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}& \text{(2)}\\
\ce{Fe^{3+} + $3\cdot$ e- &-> Fe}& E_3 &= E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}}& \text{(3)}\\
\end{aligned}
Now you are looking for the standard potential $E^\ominus(\ce{Fe/Fe^{3+}})$, so you have to consider the potential $E_3$, which can be build from $E_1$ and $E_2$:
$$E_3 = a\cdot E_1 + b\cdot E_2$$
Substituting $(1)$, $(2)$ and $(3)$ yields:
$$\begin{multline}
E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} = \\
a\cdot\left(E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) \\+
b\cdot\left(E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right)
\end{multline}$$
In order to obtain the standard potential the logarithmic terms have to cancel. Therfore we can split these equations:
\begin{aligned}
&&E^\ominus(\ce{Fe/Fe^{3+}}) &=
a\cdot E^\ominus(\ce{Fe/Fe^{2+}}) +
b\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) \\
&&\frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &=
a\cdot\left(\frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) +
b\cdot\left(\frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right)\\
\therefore&& \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &=
\frac{3a}{2} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}} +
3b \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}
\end{aligned}
In order for the $\ce{Fe^{2+}}$ terms to cancel it is fairly obvious, that $$a=\frac23,\\ b=\frac13,$$ therfore resulting in
\begin{aligned}
&&E^\ominus(\ce{Fe/Fe^{3+}}) &=
\frac{2}{3}\cdot E^\ominus(\ce{Fe/Fe^{2+}}) +
\frac{1}{3}\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}).
\end{aligned}
You can also check that against experimental data (from wikipedia):
\begin{array}{llrlr}\hline
\text{halfreaction} &&&& \text{potential}\\\hline
\ce{Fe^{2+} + $2\cdot$ e- &-> Fe} & E_1 &=& -0.44\mathrm{V} \\
\ce{Fe^{3+} + $1\cdot$ e- &-> Fe^{2+}}& E_2 &=& 0.77\mathrm{V}\\
\ce{Fe^{3+} + $3\cdot$ e- &-> Fe} & E_3 &=& -0.04\mathrm{V}\\\hline
\end{array}