In the Born approximation we have
kB(,) = kB(k,k') = [2/(424)]d3r' exp(-iqr')U(r')2,
where q = k' - k, k = v0/h, k' = v0/ (k'/k'), and  is the reduced mass.
Here  = me.
k is the incident wave vector, and (k'/k') is a unit vector pointing in the direction (,).
With v(r) = v(r) = (-Ze2/r)exp(-r/a) we have
d3r' exp(-iqr')V(r') = -Ze2r'dr'sin'd'd'exp(-r'/a)exp(-iqr'cos')
= -Ze220r'dr'exp(-r'/a)-11exp(-iqr'cos')dcos'
= -(Ze22/q)0dr'exp(-r'/a)-qr'qr'exp(-ix)dx
= -(Ze22/q)0dr'exp(-r'/a)[-i(eiqr' - e-iqr')]
= -(Ze24/q)0dr'exp(-r'/a)sin(qr') = -(Ze24/q2)q2a2/(1 + q2a2).
Therefore kB(,) = [42Z2e4/4][1/((1/a2) + q2)]2. 
With  q = 2ksin(/2) we have kB(,) = [42Z2e4/4][1/((1/a2) + 4k2sin2(/2))]2.
(b)  Let a --> , then  v(r) = (-Ze2/r)exp(-r/a) -->  v0(r) = -Ze2/r.
Then  kB(,) = [42Z2e4/4]/(16k4sin4(/2)) = [Z2e4]/(16E2sin4(/2)).
This is the Rutherford cross section.
(c)  Ratio: kB(,)part a/kB(,)part b = (q2a2/(1 + q2a2))2 = (x2/(1 + x2))2 with x = qa.
As x --> 0, the ratio approaches 0.
As a --> 0, x --> 0, the nuclear charge is completely screened, we have a neutral object.
As x --> , the ratio approaches 1.
As a --> , x --> , the screening vanishes, we have a bare nucleus.