The difference in the oxidation numbers of the two types of sulphur atoms in ${\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6$ is:

The center Sulphur atoms in $\left({\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6\right)$ have an oxidation number of 0, whereas the terminal Sulphur atoms in the compound have an oxidation number of +5.

The two different types of Sulphur atoms have oxidation values is,

 = 5 – 0

= 0

The compound's structure is provided below:

So, option A is correct.