Given that,
$$E^{0}_{O_2/H_2O} = +1.23$$ V;
$$E^{0}_{S_2O_8^{2-}/SO_4^{2-}} = 2.05$$ V;
$$E^{0}_{Br_2/Br^{-}} = +1.09$$ V;
$$E^{0}_{Au^{3+}/Au} = 1.4$$ V

The strongest oxidizing agent is:

First, recall the fundamental relation between standard reduction potential and oxidising strength.

Statement of principle: “The larger (more positive) the standard reduction potential $$E^{0}$$ of a half-cell, the greater is the tendency of the species on the left side of the reduction half-reaction to accept electrons; therefore, that species is the stronger oxidising agent.”

For each species in the question the relevant reduction half-reaction, written in the conventional “reduction” direction (electrons on the left), is as follows. We shall write the value of $$E^{0}$$ alongside each equation.

$$O_2 + 4H^{+} + 4e^{-} \longrightarrow 2H_2O \qquad E^{0}_{O_2/H_2O} = +1.23\ \text{V}$$

$$S_2O_8^{2-} + 2e^{-} \longrightarrow 2SO_4^{2-} \qquad E^{0}_{S_2O_8^{2-}/SO_4^{2-}} = +2.05\ \text{V}$$

$$Br_2 + 2e^{-} \longrightarrow 2Br^{-} \qquad E^{0}_{Br_2/Br^{-}} = +1.09\ \text{V}$$

$$Au^{3+} + 3e^{-} \longrightarrow Au \qquad E^{0}_{Au^{3+}/Au} = +1.40\ \text{V}$$

Now we compare the numerical values:

$$\begin{aligned} E^{0}_{S_2O_8^{2-}/SO_4^{2-}} &= +2.05\ \text{V} \\ E^{0}_{Au^{3+}/Au} &= +1.40\ \text{V} \\ E^{0}_{O_2/H_2O} &= +1.23\ \text{V} \\ E^{0}_{Br_2/Br^{-}} &= +1.09\ \text{V} \end{aligned}$$

Because $$+2.05\ \text{V}$$ is the highest (most positive) among all the listed potentials, the species on the left of that particular half-reaction, namely $$S_2O_8^{2-}$$ (peroxodisulphate ion), possesses the greatest tendency to be reduced.

Hence, according to the principle stated earlier, $$S_2O_8^{2-}$$ is the strongest oxidising agent in the given set.

Therefore, the option corresponding to $$S_2O_8^{2-}$$ is correct.

Hence, the correct answer is Option B.