Correct Answer: $104^\circ$

Solution :
Given: $\angle BOC=142^\circ$
In $\triangle BOC$,
⇒ $\angle BOC+\angle OBC+\angle OCB=180^\circ$
($\because$ OB and OC bisect $\angle B$ and $\angle C$ respectively)
⇒ $\angle BOC+\frac{1}{2}\angle B+\frac{1}{2}\angle C=180^\circ$
⇒ $\angle BOC=180^\circ-\frac{1}{2}(\angle B+\angle C)$
⇒ $\angle BOC=180^\circ-\frac{1}{2}(180^\circ-\angle A)$
⇒ $\angle BOC=90^\circ+\frac{1}{2}\angle A$
⇒ $142^\circ=90^\circ+\frac{1}{2}\angle A$
⇒ $52^\circ=\frac{1}{2}\angle A$
⇒ $\angle A=104^\circ$
Hence, the correct answer is $104^\circ$.