= Solution
In the <first Born approximation>, with incident and outgoing wave vectors $\mathbf k,\mathbf k'$, put $\mathbf q=\mathbf k'-\mathbf k$. For elastic scattering,
$$
q=2k\sin\frac\theta2,
\qquad f_B=-\frac{m}{2\pi\hbar^2}\int e^{-i\mathbf q\cdot\mathbf r}V(r)\,d^3r.
$$
Choose the polar axis along $\mathbf q$. The angular integral is
$$
\int e^{-iqr\cos\vartheta}\,d\Omega
=2\pi\int_{-1}^1e^{-iqru}\,du=4\pi\frac{\sin(qr)}{qr}.
$$
The radial <Dirac delta distribution> then gives
$$
f_B=-\frac{2mV_0}{\hbar^2}\int_0^\infty r^2\delta(r-R)\frac{\sin(qr)}{qr}\,dr.
$$
Thus, introducing the <dimensionless> shell strength $g=2mV_0R/\hbar^2$,
$$
\boxed{f_B(\theta)=-\frac{2mV_0R}{\hbar^2q}\sin(qR)
=-gR\frac{\sin(qR)}{qR}}.
$$
The forward value is the continuous limit $f_B(0)=-gR$. This is the <delta-shell Born amplitude>, with its angular dependence expressed through the <momentum transfer>.
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