= Solution
For the conventional <body-centred cubic lattice>, the <crystal basis structure factor> of identical atoms is
$$
F_{hkl}=f_{\mathrm W}\bigl[1+e^{i\pi(h+k+l)}\bigr].
$$
It vanishes for odd $h+k+l$ and equals $2f_{\mathrm W}$ for even $h+k+l$. Thus the first <powder diffraction> ring is the $\{110\}$ family, with <interplanar spacing> $d_{110}=a/\sqrt2$.
The measured deflection is the angle between incident and scattered rays, $2\theta=40.26^\circ$, so the <Bragg angle> is $\theta=20.13^\circ$. Using first-order <Bragg diffraction>, $2d\sin\theta=\lambda$, gives
$$
\boxed{a=\frac{\lambda}{\sqrt2\sin\theta}=0.31683\,\mathrm{nm}.}
$$
The <mass density> follows from two atoms per conventional <unit cell>, with molar mass $M=0.1838\,\mathrm{kg\,mol^{-1}}$ and <Avogadro constant> $N_A$:
$$
\boxed{\rho=\frac{2M}{N_Aa^3}=1.9194\times10^4\,\mathrm{kg\,m^{-3}}\simeq19.19\,\mathrm{g\,cm^{-3}}.}
$$
Using $40.26^\circ$ as the <Bragg angle> would confuse $\theta$ with $2\theta$ and give an incorrect <lattice constant> and <mass density>.
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