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I've worked this problem and seem to be off by a factor of $2$ somehow. From Callister, 6th edition, problem 3.56W (but I don't have access to the "W" web material that actually explains Bragg diffraction). Using other texts, I found that plane spacing $d$ satisfies $$\frac{1}{d^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2},$$ which for the $(311)$ plane in an FCC lattice (therefore cubic) gives $a = \sqrt{11}d$.
The problem gives $\theta = 36.12^\circ$ and $\lambda = \pu{0.0711 mm}$. Bragg's Law says $$n\lambda = 2 d \sin \theta,$$ so for first order diffraction (the problem says reflection but I assume that is a mistake) $n=1$ and I get $d = \pu{0.0603 mm}$, which gives $a = \pu{0.2000 mm}$ and using FCC, $r = \pu{0.0707 nm}$, which is about half the correct answer according to Wikipedia $(\pu{142\pm 7 pm}$).

Can anyone tell me where I went wrong?

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Nobody answered this and I figured it out, so I'll explain here.

It turns out that glancing angle is $\theta$, the angle the incoming beam makes with the surface (not the normal to the surface, as is usual in optics).

Meanwhile the diffraction angle is $2\theta$, the total amount the incoming beam gets "turned" to become the outgoing beam.

The problem stated the diffraction angles. They need to be divided by $2$ before being plugged into the Bragg Law.

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