A pure metal M crystallizes in a face-centered cubic (fcc) lattice....

In a face-centered cubic (fcc) lattice, the number of tetrahedral voids is twice the number of atoms.

Let the number of atoms in the sample be $N$.

Number of tetrahedral voids $= 2N = 3.612 \times 10^{24}$

$\Rightarrow N = \dfrac{3.612 \times 10^{24}}{2} = 1.806 \times 10^{24}$ atoms

Number of moles of metal $\text{M} = \dfrac{N}{N_A} = \dfrac{1.806 \times 10^{24}}{6.02 \times 10^{23}} = 3 \text{ mol}$

Mass of the sample = moles $\times$ atomic mass $= 3 \times 27 = 81 \text{ g}$

Answer: $81$