Answer
$9\times 10^{-13}\;\rm N$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the plane bisects the dipole which means that the plane is perpendicular to the dipole. The author told us that the proton is on this plane that bisects the dipole.
Thus, the electric field exerted by the molecule dipole on the position of the proton is given by
$$E_{dipole}= \dfrac{1}{4\pi \epsilon_0}\dfrac{p}{r^3} $$
Hence, the force exerted by the molecule dipole on the proton is given by
$$F=q_{\rm proton}E_{dipole}= \dfrac{1}{4\pi \epsilon_0}\dfrac{q_{\rm proton}p}{r^3}$$
Plug the known;
$$F = \dfrac{1}{4\pi (8.85\times 10^{-12})}\dfrac{(1.6\times 10^{-19})(5\times 10^{-30})}{(2\times 10^{-9})^3}$$
$$F=\color{red}{\bf 9\times 10^{-13}}\;\rm N\tag{in $-\vec p$ direction}$$
The direction of this force is just opposite to the dipole moment direction. See the figure below.
