The major product [C] of the following reaction sequence will be:

The starting compound is ethanal, $$CH_3-CHO$$.

Case 1: Formation of the aldol (compound B)
Ethanal possesses at least one $$\alpha$$-hydrogen. In the presence of a dilute aqueous base (usually $$NaOH$$ or $$KOH$$, 273-278 K), one molecule of ethanal is enolised and the enolate ion thus formed attacks the carbonyl carbon of a second molecule of ethanal. This is the classic aldol addition.

The overall transformation is
$$2\,CH_3-CHO \;\xrightarrow[\;273\text{-}278\text{ K}\;]{\;dil.\;NaOH\;} CH_3-CH(OH)-CH_2-CHO$$

The product B is 3-hydroxybutanal, a β-hydroxy aldehyde (aldol).

Case 2: Dehydration of the aldol (compound C)
When the reaction mixture is warmed (or simply allowed to stand for a longer time under the same basic conditions), the β-hydroxy aldehyde undergoes base-induced dehydration (loss of water) to give an α,β-unsaturated aldehyde.

The elimination occurs between the β-hydroxyl group and an α-hydrogen:
$$CH_3-CH(OH)-CH_2-CHO \;\xrightarrow[\;\text{Δ}\;]{\;dil.\;NaOH\;} CH_3-CH=CH-CHO + H_2O$$

The product C is crotonaldehyde (but-2-enal). Experimentally, the trans (E) stereoisomer predominates because it is more stable.

Hence the major product C is the α,β-unsaturated aldehyde $$CH_3-CH=CH-CHO$$, represented in the options as Option A.

Final answer: Option A which is: trans-but-2-enal (crotonaldehyde).