Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that
$$p=\alpha E$$
and we need to find the units of $\alpha$, so
$$\alpha=\dfrac{p}{E}$$
We know that the unit of $p$ is $\rm C\cdot m$ and that the unit of $E$ is $\rm N/C$
Thus,
$$\rm \alpha-Unit=\dfrac{C\cdot m}{N/C}=\dfrac{C^2\cdot m}{N}=\dfrac{C^2\cdot m\cdot s^2}{kg\cdot m} $$
$$\boxed{\rm \alpha-Unit =\dfrac{C^2\cdot s^2}{kg } }$$
$$\color{blue}{\bf [b]}$$
According to Newtin's third law, the force exerted by the dipole on the ion is equal in magnitude but opposite in direction to the force exerted by the ion on the dipole.
To find $F_{\text{ion on dipole}}$, we need to find the electric field exerted on the dipole by the ion $q$,
$$E_{\text{ion on dipole}}=\dfrac{kq}{r^2}\;\hat i\tag 1$$
Hence, the dipole moment is
$$p=\alpha E_{\text{ion on dipole}}$$
Plug from (1),
$$p= \dfrac{\alpha kq}{r^2}\;\hat i\tag 2$$
Now we need to find the electric field exerted by the dipole on the ion, which is on the axis,
$$E_{\text{dipole on ion}}=\dfrac{k (2p)}{r^3}$$
Plug from (2),
$$E_{\text{dipole on ion}}=\dfrac{2k }{r^3} \dfrac{\alpha kq}{r^2}\;\hat i$$
$$E_{\text{dipole on ion}}= \dfrac{2\alpha k^2q}{r^5}\;\hat i$$
Hence, the force exerted by the dipole on the ion is given by
$$F_{\text{dipole on ion}}=qE_{\text{dipole on ion}}=\dfrac{2\alpha k^2q^2}{r^5}\;\hat i$$
Therefore,
$$F_{\text{ion on dipole}}=-\dfrac{2\alpha k^2q^2}{r^5}\;\hat i$$
Recalling that $k=1/4\pi \epsilon_0$
$$\boxed{F_{\text{ion on dipole}}=\dfrac{-1}{(4\pi \epsilon_0)^2}\dfrac{2\alpha q^2}{r^5}\;\hat i}$$
