The structure of A in the given reaction is

The reaction shown is an $$\mathrm{\alpha}$$-alkylation of a methyl ketone.

The starting ketone is:

$$\mathrm{R-CO-CH_3}$$

In the presence of sodium hydroxide:

$$\mathrm{NaOH}$$

one of the acidic $$\mathrm{\alpha}$$-hydrogens is removed to form a resonance-stabilised enolate ion.

$$\mathrm{R-CO-CH_3 + OH^- \longrightarrow R-CO-CH_2^- + H_2O}$$

The enolate ion then reacts with ethyl bromide:

$$\mathrm{CH_3CH_2Br}$$

through an $$\mathrm{S_N2}$$ substitution reaction.

The negatively charged $$\mathrm{\alpha}$$-carbon attacks the ethyl group and displaces bromide ion.

$$\mathrm{R-CO-CH_2^- + CH_3CH_2Br \longrightarrow R-CO-CH_2CH_2CH_3 + Br^-}$$

Thus, the original methyl group $$\mathrm{(-CH_3)}$$ is converted into a propyl chain:

$$\mathrm{(-CH_2CH_2CH_3)}$$

attached to the carbonyl carbon.

Therefore, the correct product is:

$$\boxed{\mathrm{Structure\ C}}$$