The major product of the following reaction is :

$\mathrm{NaOH} \text{and} {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$ is alcoholic$\mathrm{NaOH}$ which is used for dehydrohalogenation of alkenes. This beta elimination is anti-elimination and Saytzeff alkene (more stable) is major product. In the given reaction, more stable conjugated alkene is formed as major product.

Note : There is ambiguity in answer given by NTA which is $4$.