Solution:
$$
\begin{aligned}
&\text { Given: } \vec{F}=3 \hat{j} N, \\
&\vec{r}=2 \hat{k} \\
&\text { We know, } \vec{\tau}=\vec{r} \times \vec{F}=2 \hat{k} \times 3 \hat{j} \\
&=6(\hat{k} \times \hat{j})=6(-\hat{i}) \\
&\vec{\tau}=-6 \hat{i} N m
\end{aligned}
$$