Answer
$\tau=-(1.98*10^{-26}N*m)\hat{k}$
Work Step by Step
We are given the vectors $\vec{P}$ and $\vec{E}$, so we can use the equation $\tau=\vec{P}\times\vec{E}$
$\vec{P}=(3.00\hat{i}+4.00\hat{j})(1.24*10^{-30}C*m)$ and $\vec{E}=4000 \frac{N}{C}$
So, we create a matrix (for the sake of simplicity, let $1.24*10^{-30}=x$):
\[\begin{bmatrix}\hat{i}&\hat{j}&\hat{k} \\3.00x&4.00x&0\\4000 & 0 & 0\end{bmatrix}\]
Taking the cross product of this matrix, we get $\tau=-(1.98*10^{-26}N*m)\hat{k}$
