= Solution
In the <first Born approximation>, let $q=|\mathbf k'-\mathbf k|=2k\sin(\theta/2)$ be the transferred wave number. The angular <integral> in the three-dimensional <Fourier transform> is $4\pi\sin(qr)/(qr)$, so
$$
f_B(\theta)=-\frac{m}{2\pi\hbar^2}\int e^{-i\mathbf q\cdot\mathbf r}V_0\delta(r-R)\,d^3r
=-\frac{2mV_0R^2}{\hbar^2}\frac{\sin(qR)}{qR}.
$$
Writing $g=2mV_0R/\hbar^2$ and $\operatorname{sinc}u=\sin u/u$, with value one at zero, the <delta-shell Born amplitude> gives
$$
\boxed{f_B=-gR\operatorname{sinc}[2kR\sin(\theta/2)],\qquad
\frac{d\sigma_B}{d\Omega}=g^2R^2\operatorname{sinc}^2[2kR\sin(\theta/2)].}
$$
For $kR\ll1$ the scattering is approximately isotropic, with total <cross section> $4\pi g^2R^2$. For $kR\gg1$ it has a forward lobe of angular width of order $(kR)^{-1}$ and oscillatory zeros where $2kR\sin(\theta/2)=n\pi$. At a fixed nonzero <scattering angle>, its envelope decreases as $k^{-2}$. The right-hand panel of the figure shows this angular dependence.
For real $V_0$ the first-order <scattering amplitude> is real, so inserting it literally in the <optical theorem> would predict zero total <cross section>, contrary to its nonzero squared <modulus>. \b[The first Born amplitude alone does not satisfy the optical theorem at order $V_0^2$.] This is an order-of-approximation issue: the squared first-order amplitude is already second order, and the second Born term supplies the required second-order imaginary forward amplitude. At first order both sides vanish. A weak-shell regime is needed for the Born truncation itself to be accurate.
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