Certainly late for your homework problem, but figure below shows the schematic of elastic scattering. The scattering angle is $\Theta$. The momentum transfer vector is $\Delta \bf{p}$.
Since your question starts with potential, we shall obtain the central force first:
\begin{align}
F(r) &= \frac{dV}{dr} \\
&= -\frac{\alpha}{r^2}
\end{align}
Since the force is central, at any time the force in the direction of $\Delta \bf{p}$ is $F \cos\phi$, so the total momentum transfer is
\begin{align}
\int_{-\infty}^{\infty} F \cos{\phi}\,dt
\end{align}
Using one of your information $bv_0=r^2\frac{d\phi}{dt}$ (which is obtainable from conservation of angular momentum) we make the change of variable
\begin{align}
dt = \frac{r^2}{bv_0}d\phi
\end{align}
. In the limit $t\rightarrow \pm \infty$, the angle $\phi$ approaches $\pm(\pi-\Theta)/2$ as measured from the direction of momentum transfer axis. The integration becomes,
\begin{align}
\int_{-\frac{\pi-\Theta}{2}}^{\frac{\pi-\Theta}{2}} \left(-\frac{\alpha}{r^2}\right)\cos{\phi}\,\frac{r^2}{bv_0}d\phi &= \frac{\alpha}{bv_0}\left[\sin\phi\right]^{\frac{\pi-\Theta}{2}}_{-\frac{\pi-\Theta}{2}} \\
&= \frac{2\alpha}{bv_0}\cos{\left(\frac{\Theta}{2}\right)}
\end{align}
$N(\theta)/N_i$ is a probability density not a probability. To get a probability you have to integrate it over $\theta$ or maybe multiply it by $d\theta$. That makes the numbers reasonable again.
Usually the difference is clear because of dimensions, if you're dealing with distributions in things like time or position or energy. But angles are dimensionless that doesn't help here.
Best Answer
In the Wikipedia article about rutherford scattering the derivation of the scattering cross section $$ \frac{d\sigma}{d\Omega} =\left(\frac{ Z_1 Z_2 e^2}{8\pi\epsilon_0 m v_0^2}\right)^2 \csc^4{\left(\frac{\Theta}{2}\right)}$$ is given. Let's rewrite that in your notation: $Z_1 = Z$, $Z_2 = 4$, $k = \frac{1}{4\pi\varepsilon_0}$ and $KE = \frac{1}{2}m v^2$: $$ \frac{d\sigma}{d\Omega} =\left(\frac{ Z e^2 k}{KE}\right)^2 \csc^4{\left(\frac{\Theta}{2}\right)} = \frac{Z^2 k^2 e^4}{ KE^2\sin^4(\theta/2)}$$ The relation between the cross section and the number of detected particles is $$\frac{N(\theta)}{N_i} = \frac{nL}{4r^2} \frac{d\sigma}{d\Omega}$$ $nL$ gives the number of targets to scatter from and $1/(4r^2)$ is because at a bigger distance the intensity goes down by this factor (I'm sure about the $r^2$ but not exactly about the factor. I thought it would be $2\pi$ so maybe I did an error there.)
I hope this helps.