Oxidation states of iron in Roussin's salts
I suppose it's asked about formal oxidation numbers of iron as there is not enough information to make assumptions about the ones that reflect electronic structure.
For instance, analysis of interatomic distances could point to a delocalized electron pair, but you are not given any linear or angular geometrical parameters to begin with.

It's only the symmetry that's evident from the crystal structure: the thioferrate cluster $\ce{[Fe4S3(NO)7]-}$ belongs to $C_\mathrm{3v}$ point group, hence the only possible arrangement of four iron atoms is a tetrahedral one.
Another conclusion that can be made is that three equatorial iron atoms are equivalent ($\ce{Fe(B)}$, $\ce{Fe(C)}$, $\ce{Fe(D)}$), and the last apical one ($\ce{Fe(A)}$) most likely deviates in terms of oxidation state.

Supposing formal oxidation number ($\text{O.N.}$) of three equivalent irons and the last one are $y$ and $x$, respectively, and the fact that the average $\text{O.N.}$ is $-0.5$, the following holds true:

$$x + 3y = -0.5\cdot 4 \implies y = -\frac{x + 2}{3}$$

As formal oxidation numbers are integers, there are several pairs giving the solution:

$$
\begin{array}{l|rrrrr}
\hline
x &amp; \ldots &amp; -5 &amp; -2 &amp; \color{green}{1} &amp; 4 &amp; 7 &amp; \ldots \\
\hline
y &amp; \ldots &amp; 1 &amp; 0 &amp; \color{green}{-1} &amp; -2 &amp; -3 &amp; \ldots\\
\hline
\end{array}
$$

among which the least "extreme" (e.g. the one with the absolute lowest negative $\text{O.N.}$) for a metal would be the $-1;+1$ pair; so that for $\overset{0}{\ce{Fe}}:~[\ce{Ar}]\,(\mathrm{3d})^6\,(\mathrm{4s})^2$ your answer gives two configurations, $(\mathrm{3d})^7$ for $\overset{+1}{\ce{Fe}}\ce{(A)}$ and $(\mathrm{3d})^9$ for $\overset{-1}{\ce{Fe}}\ce{(B,C,D)}$.