Among the oxides of nitrogen: N$$_2$$O$$_3$$, N$$_2$$O$$_4$$ and N$$_2$$O$$_5$$; the molecule(s) having nitrogen-nitrogen bond is/are:

First, let us recall that the existence of a nitrogen-nitrogen bond can be confirmed only by looking at the actual Lewis or structural formula of the oxide. A mere count of the atoms in the molecular formula is not sufficient; we must see whether the two nitrogen atoms are directly joined to each other or whether they are separated by some other atom such as oxygen.

We examine the three oxides one by one.

For dinitrogen trioxide, $$\mathrm{N_2O_3}$$, the experimentally established structure is written as $$\mathrm{O_2N{-}NO}$$. Here the two nitrogen atoms are directly connected by a single bond, that is, there is an explicit $$\mathrm{N{-}N}$$ linkage. Thus $$\mathrm{N_2O_3}$$ definitely possesses a nitrogen-nitrogen bond.

Next we inspect dinitrogen tetroxide, $$\mathrm{N_2O_4}$$. This molecule is obtained by dimerisation of nitrogen dioxide $$\mathrm{NO_2}$$, and its structure is $$\mathrm{O_2N{-}NO_2}$$. Once again the two nitrogen centers are directly attached to each other, giving a clear $$\mathrm{N{-}N}$$ bond in the molecule. Therefore $$\mathrm{N_2O_4}$$ also contains a nitrogen-nitrogen bond.

Finally we look at dinitrogen pentoxide, $$\mathrm{N_2O_5}$$. The accepted Lewis structure for this oxide is $$\mathrm{O_2N{-}O{-}NO_2}$$. In this arrangement the two nitrogen atoms are not bonded to each other; instead, an oxygen atom sits between them, forming an $$\mathrm{N{-}O{-}N}$$ bridge. As a result there is no direct $$\mathrm{N{-}N}$$ bond in $$\mathrm{N_2O_5}$$.

Putting these observations together, we have found a direct nitrogen-nitrogen bond in $$\mathrm{N_2O_3}$$ and in $$\mathrm{N_2O_4}$$, but not in $$\mathrm{N_2O_5}$$.

Hence, the correct answer is Option A.