Rearranging this equation, we get:
$$d = \frac{n\lambda}{2\sin\theta}$$
Plugging in the values given in the problem, we get:
$$d = \frac{(1)(0.090\ \mathrm{nm})}{2\sin(15.2^\circ)} \approx 0.166\ \mathrm{nm}$$
To convert this to picometers, we multiply by
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