= Solution
Translate the scattering solution for the atom at the origin by $\mathbf a$. Its incident wave becomes $e^{i\mathbf k\cdot(\mathbf r-\mathbf a)}$; multiplying the whole solution by $e^{i\mathbf k\cdot\mathbf a}$ restores the prescribed incident wave. At large $r$,
$$
|\mathbf r-\mathbf a|=r-\widehat{\mathbf r}\cdot\mathbf a+O(r^{-1}),
\qquad\widehat{\mathbf r-\mathbf a}=\widehat{\mathbf r}+O(r^{-1}).
$$
Thus the outgoing amplitude acquires the phase $e^{i(\mathbf k-\mathbf k')\cdot\mathbf a}$, with $\mathbf k'=k\widehat{\mathbf r}$.
In the single-scattering approximation, amplitudes from lattice sites add coherently. If $\mathbf Q=\mathbf k-\mathbf k'$,
$$
\frac{d\sigma}{d\Omega}=|f(\widehat{\mathbf r})|^2
\left|\sum_{\mathbf a\in\mathcal L_N}e^{i\mathbf Q\cdot\mathbf a}\right|^2,
\qquad
\boxed{\Delta(\mathbf Q)=\frac1N
\left|\sum_{\mathbf a\in\mathcal L_N}e^{i\mathbf Q\cdot\mathbf a}\right|^2.}
$$
For a rectangular block of a <Bravais lattice> with $N_j$ sites along primitive vectors $\mathbf a_j$ and $N=N_1N_2N_3$, the <geometric series> gives
$$
\boxed{\Delta(\mathbf Q)=\frac1N
\prod_{j=1}^3
\left[\frac{\sin(N_j\mathbf Q\cdot\mathbf a_j/2)}
{\sin(\mathbf Q\cdot\mathbf a_j/2)}\right]^2.}
$$
Each ratio has its limiting value $N_j$ when its argument is a multiple of $2\pi$. Thus the peaks have height $N$ and widths of order $1/N_j$ in those phase variables, at $\mathbf Q\cdot\mathbf a_j\in2\pi\mathbb Z$, exactly the <reciprocal lattice> vectors.
With the incident-wave normalization used here, the <Born approximation> is
$$
f_B(\widehat{\mathbf r})=-\frac{m_e}{2\pi\hbar^2}
\int_{\mathbb R^3}e^{i(\mathbf k-\mathbf k')\cdot\mathbf r}
V(\mathbf r)\,d^3r.
$$
For the stated attractive three-dimensional delta potential,
$$
\boxed{f_B=\frac{m_ea}{2\pi\hbar^2},}
$$
independent of scattering angle. This is the formal first Born amplitude, not a claim that an unregularized three-dimensional delta interaction defines an exact scattering Hamiltonian.
The 60-degree scattering angle is twice the Bragg angle, so $\theta=30^\circ$. First-order <elastic Bragg scattering condition> gives $2d\sin\theta=\lambda$, hence
$$
\boxed{d=\lambda.}
$$
Higher orders would give $d=n\lambda$; the smallest, usual first-order spacing is the likely answer.
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