This arrangement of electric dipole in electric field will undergo SHM, when the rod is set free. Time period of this SHM is given by
$$
T=2 \pi \sqrt{\frac{I}{\mathrm{PE}}}
$$
where, $I=$ moment of inertia of dipole, $p=$ electric dipole moment and $E=$ electric field. $I=2 m r^2$
$$
\begin{aligned}
& \text { mass, } m=9.8 \times 10^{-3} \mathrm{~kg}, r=\frac{50}{2} \times 10^{-2} \\
& I=2 \times 9.8 \times 10^{-3} \times\left(\frac{50}{2} \times 10^{-2}\right)^2 \\
& I=1225 \times 10^{-4} \mathrm{k}-g^2 m^{-2}=10^{-5} \mathrm{C}-\mathrm{m} \\
& p=q(2 l)=20 \times 10^{-6} \times 50 \times 10^{-2} \\
& E=121 \mathrm{~N} / \mathrm{C} \\
& T=2 \pi \sqrt{\frac{1225 \times 10^{-4}}{10^{-5} \times 121}}=2 \times 3.14 \times 3.181 \\
& T=19.98 \cong 20 \mathrm{~s}
\end{aligned}
$$

But time taken by the rod to become parallel will be $T / 4$, so required time is $5 \mathrm{~s}$.