The correct option is(A): 150 J.
\(\overrightarrow{F}=(-2\widehat{i}+15\widehat{j}+6\widehat{k})\)N and distance (d) = 10\(\widehat{j}\, m\).
Work done
W=\(\overrightarrow{F}\cdot \overrightarrow{d}= (-2\widehat{i}+15\widehat{j}+6\widehat{k})\cdot (10\widehat{j})=150 N-m =150\, J\).