= Solution
Define $A=mg^2/(2\pi\hbar^2)$, which has units of <energy> <times> <length>. Substituting the result of (b) into (c) gives
$$
\left(E_k-\epsilon_0+\frac A\delta+iAk\right)d_k=g.
$$
Absorb the real cutoff shift into the physical detuning,
$$
\widetilde\epsilon_0=\epsilon_0-\frac A\delta,\qquad
d_k=\frac g{E_k-\widetilde\epsilon_0+iAk}.
$$
It follows that
$$
f_k=-\frac A{E_k-\widetilde\epsilon_0+iAk}.
$$
Using $k=\sqrt{2mE_k}/\hbar$, identify
$$
\boxed{\gamma=\frac{\sqrt{2m}}\hbar A
=\frac{m^{3/2}g^2}{\sqrt2\,\pi\hbar^3},\qquad
f_k=-\frac{\hbar\gamma}{\sqrt{2m}}\,
\frac1{E_k-\widetilde\epsilon_0+i\gamma\sqrt{E_k}}.}
$$
Here $\gamma$ has units of square root of <energy>. This is the small-cutoff, renormalized resonant-level amplitude; terms vanishing with $k\delta$ have been omitted. Taking $\delta\to0$ at fixed measured detuning requires the bare $\epsilon_0$ to depend on the cutoff.
The sign is checked by $f_k^{-1}=-(E_k-\widetilde\epsilon_0)/A-ik$, which has the elastic $s$-wave <unitarity> form. Thus $\operatorname{Im}f_k=k|f_k|^2$, consistent with the <optical theorem> and $\sigma_{\rm tot}=4\pi|f_k|^2$. If the renormalized level is positive, at $E_k=\widetilde\epsilon_0$ the amplitude is $i/k$ and the <scattering cross-section> reaches $4\pi/k^2$.
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