= Solution
The <reciprocal lattice> is $\Lambda^*=\{G:G\cdot\ell\in2\pi\mathbb Z\text{ for every }\ell\in\Lambda\}$. With oriented primitive-cell volume $v_c=a_1\cdot(a_2\times a_3)$, its basis is $b_1=2\pi(a_2\times a_3)/v_c$ and the cyclic variants; thus $b_i\cdot a_j=2\pi\delta_{ij}$.
The total potential is $V(x)=\sum_\ell U(x-\ell)$. Using the <Fourier transform> convention $\widetilde U(q)=\int e^{-iq\cdot x}U(x)d^3x$, translation gives $\widetilde V(q)=\widetilde U(q)\Delta(q)$, with
$$
\Delta(q)=\prod_{j=1}^3\sum_{n=0}^{N_j-1}e^{-in q\cdot a_j}.
$$
The given <Born approximation> therefore supplies the amplitude $\boxed{-m\Delta(q)\widetilde U(q)/(2\pi\hbar^2)}$. A finite <geometric series> gives, when the denominators are nonzero,
$$
\boxed{|\Delta(q)|=\prod_{j=1}^3\left|\frac{\sin(N_jq\cdot a_j/2)}{\sin(q\cdot a_j/2)}\right|.}
$$
At reciprocal-lattice vectors each phase is one, so $|\Delta|=N_1N_2N_3$, the number of atoms. For large finite crystals these narrow peaks represent coherent <Bragg scattering>, with peak intensity proportional to the squared number of atoms; their finite width reflects the finite crystal size.
For the specified <face-centered cubic lattice>, $v_c=a^3/4$ and
$$
b_1=\frac{2\pi}a(-\hat x+\hat y+\hat z),\quad b_2=\frac{2\pi}a(\hat x-\hat y+\hat z),\quad b_3=\frac{2\pi}a(\hat x+\hat y-\hat z).
$$
The <reciprocal lattice> have coordinates $(2\pi/a)(h,k,l)$ with $h,k,l$ all even or all odd. The two shortest nonzero length families are $G_1=2\pi\sqrt3/a$ for $(111)$ and $G_2=4\pi/a$ for $(200)$. Elastic scattering satisfies $|q|=2|k|\sin(\theta/2)$. The assumed incident magnitude exceeds $G_2/2$, so both families are kinematically accessible, and
$$
\boxed{\frac{\sin(\theta_1/2)}{\sin(\theta_2/2)}=\frac{G_1}{G_2}=\frac{\sqrt3}{2}.}
$$
This is the angular prediction for suitable crystal orientations, or for a powder containing those orientations. A fixed single-crystal orientation must additionally satisfy the vector <Elastic Bragg scattering condition> $2k\cdot G+|G|^2=0$; the magnitude bound alone cannot guarantee both peaks for every incident direction.
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