A metal M forms hexagonal close-packed structure. The total number of voids in $$0.02$$ mol of it is ______ $$\times 10^{21}$$ (Nearest integer)
(Given N$$_A = 6.02 \times 10^{23}$$)

In a hexagonal close-packed (hcp) structure, for every atom there are 2 tetrahedral voids and 1 octahedral void, giving a total of 3 voids per atom.

The number of atoms in 0.02 mol is $$N = 0.02 \times N_A = 0.02 \times 6.02 \times 10^{23} = 1.204 \times 10^{22}$$.

The total number of voids is $$3N = 3 \times 1.204 \times 10^{22} = 3.612 \times 10^{22} = 36.12 \times 10^{21}$$.

Rounding to the nearest integer, the total number of voids is $$\boxed{36} \times 10^{21}$$.