$(a)\;9\qquad(b)\;10\qquad(c)\;11\qquad(d)\;12$

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There are four body diagonals and atoms on the body diagonal are not shared by any other unit cell.

$\therefore$ Atoms at the corner = $8\times \large\frac{1}{8}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 1$

Atoms on the body diagonal = $2\times4$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; =8$

$\therefore$ Total number of atoms = 1+8 = 9

Hence answer is (a)

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