Answer
$0.283\;\rm rad/s$
Work Step by Step
As we see in the figures below, we need to find the angular velocity of the dipole when it rotates about 90$^\circ$ counterclockwise so it will be aligned with the electric field lines.
Since the dipole rotates on a frictionless pivot at its center, the energy is conserved.
$$E_i=E_f$$
$$K_i+U_{i,\rm dipole}=K_f+U_{f,\rm dipole}$$
where $_i\rightarrow$ is when the dipole is perpendicular to the electric field lines. And $_f\rightarrow$ when the dipole is aligned with the electric field lines.
Recall that $U_{\rm dipole}=- pE\cos\theta$
$$\frac{1}{2}I\omega_i^2+(-pE\cos90^\circ)=\frac{1}{2}I\omega_f^2+(-pE\cos0^\circ)$$
$$\frac{1}{2}I\omega_i^2 =\frac{1}{2}I\omega_f^2 -pE $$
where $\omega_i=0$, so
$$ \frac{1}{2}I\omega_f^2 =pE $$
$$ \omega_f =\sqrt{\dfrac{2pE }{I}}$$
Recall that $p=qL$, and that $I=2m(\frac{L}{2})^2$
$$ \omega_f =\sqrt{\dfrac{8qLE }{2mL^2}}$$
$$ \omega_f =\sqrt{\dfrac{8q E }{2mL }}$$
Plug the known;
$$ \omega_f =\sqrt{\dfrac{8(2\times 10^{-9})(1000) }{2(1\times 10^{-3})(0.10) }}$$
$$ \omega_f =\color{red}{\bf 0.283}\;\rm rad/s$$
