1. Write the given standard reduction steps

From the diagram:

$\mathrm{FeO}_4^{2-} \xrightarrow{+2.0\,V} \mathrm{Fe}^{3+} \xrightarrow{0.8\,V} \mathrm{Fe}^{2+} \xrightarrow{-0.5\,V} \mathrm{Fe}^0$

we interpret the first two relevant reduction half-reactions as:

$\mathrm{FeO}_4^{2-} + 8H^+ + 3e^- \rightarrow \mathrm{Fe}^{3+} + 4H_2O \qquad E^\circ = 2.0\,V$

and

$\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \qquad E^\circ = 0.8\,V$

We need $E^\circ$ for:

$\mathrm{FeO}_4^{2-} + 8H^+ + 4e^- \rightarrow \mathrm{Fe}^{2+} + 4H_2O$

2. Use the relation involving Gibbs energy

Standard electrode potentials are not added directly unless the number of electrons is same. We use:

$\Delta G^\circ = -nFE^\circ$

For the first step:

$\Delta G_1^\circ = -3F(2.0) = -6.0F$

For the second step:

$\Delta G_2^\circ = -1F(0.8) = -0.8F$

Adding them:

$\Delta G_{\text{total}}^\circ = -6.0F -0.8F = -6.8F$

3. Find the overall electrode potential

The overall reaction involves $4$ electrons, so

$\Delta G_{\text{total}}^\circ = -4F E^\circ_{\mathrm{FeO}_4^{2-}/\mathrm{Fe}^{2+}}$

Thus,

$-4F E^\circ = -6.8F$

$E^\circ = \frac{6.8}{4} = 1.7\,V$

4. Match with options

$E^\circ_{\mathrm{FeO}_4^{2-}/\mathrm{Fe}^{2+}} = 1.7\,V$

So the correct option is:

D: $1.7\,V$