The torque \( \vec{\tau} \) on a dipole in an electric field is given by:
\[
\vec{\tau} = \vec{p} \times \vec{E}
\]
Given:
\[
|\vec{p}| = 1.0 \times 10^{-12} \ \text{Cm}, \quad |\vec{E}| = 2.0 \times 10^4 \ \text{NC}^{-1}
\]
\[
\hat{p} = \hat{i}, \quad \hat{E} = \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}
\]
Now compute the cross product:
\[
\vec{\tau} = pE \sin\theta
\]
Angle \( \theta \) between \( \hat{i} \) and the vector \( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \) can be found using:
\[
\sin\theta = \left| \hat{i} \times \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) \right| = \left| \frac{\sqrt{3}}{2} (\hat{i} \times \hat{i}) + \frac{1}{2} (\hat{i} \times \hat{j}) \right| = \left| 0 + \frac{1}{2} \hat{k} \right| = \frac{1}{2}
\]
Thus:
\[
\tau = pE \sin\theta = (1.0 \times 10^{-12})(2.0 \times 10^4) \left( \frac{1}{2} \right)
= 1.0 \times 10^{-8} \ \text{Nm}
\]