= Solution
The <Fermi sphere> <volume> in reciprocal space is
$$
\boxed{V_F=\frac{4\pi}3k_F^3=46.9429\ \text{\AA}^{-3}}.
$$
The primitive real-space cell has <volume> $a^3/2$, so the first <Brillouin zone> has <volume>
$$
\boxed{V_{\rm BZ}=\frac{(2\pi)^3}{a^3/2}=15.6476\ \text{\AA}^{-3}},\qquad
\boxed{V_F=3V_{\rm BZ}}.
$$
Each band accommodates two <electrons> per primitive cell when spin is counted. Six <electrons> therefore require three such zone-volumes of occupied reciprocal-space states in the free-electron counting picture. The extended-zone sphere is not confined to the first zone; folding it produces multiple bands. It does not imply that the actual <tungsten> <Fermi surface> is spherical.
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