The major product of the reaction is:

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(C) The given reactant is an erythro-isomer. In an $E_2$ elimination reaction,the base abstracts a proton that is anti-periplanar to the leaving group

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(C) The given reactant is an erythro-isomer. In an $E_2$ elimination reaction,the base abstracts a proton that is anti-periplanar to the leaving group $(-Br)$.For the erythro-isomer,the anti-elimination leads to the formation of the $trans$-alkene as the major product.In the given structure,the $Ph$ groups are on opposite sides in the $trans$ configuration,which corresponds to the product shown in option $C$.

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