**Correct option (b) 8.16 × 10**^{12} atoms/mm^{2 }

**Explanation:**

Lead has FCC structure

The given interatomic distance = 3.499 × 10^{-10} m

∴ 2r = 3.499 × 10^{-10} m

If ‘a’ is the side of the square

∴√2a^{2} = 4r

∴ a = 4.95 × 10^{-10} m

Now, area of (010) plane = a^{2} = (4.95 × 10^{-10}) ^{2} m^{2}

For FCC, Number of atoms in (010) plane = 2

∴ number of atoms per millimeter square

= 2/(4.95 x 10^{-10})^{2}m^{2}

=^{ }8.16 × 10^{12} atoms/mm^{2}