Answer is: there are 1.82·10²⁰ silicon-30 atoms.
V₁(alloy) = 15 cm · 12.5 cm · 0.25 cm.
V₁(alloy) = 46.875 cm³; volume of the alloy without hole.
V(hole) = 0.25 cm · (1.25 cm)² · 3.14.
V(hole) = 1.23 cm³
V₂(alloy) = 46.875 cm³ - 1.23 cm³.
V₂(alloy) = 45.65 cm³; volume of the alloy with hole.
m(alloy) = V₂(alloy) · d(alloy).
m(alloy) = 45.65 cm³ · 8.80 g/cm³.
m(alloy) = 401.7 g; mass of the alloy.
ω(Si) = 0.073% ÷ 100%.
ω(Si) = 0.00073.
m(Si) = ω(Si) · m(alloy).
m(Si) = 0.00073 · 401.7 g.
m(Si) = 0.293 g; mass of silicon it the alloy.
m(Si-30) = m(Si) · ω(Si-30).
m(Si-30) = 0.293 g · 0.031.
m(Si-30) = 0.0091 g; mass of the silicon-30.
n(Si-30) = m(Si-30) ÷ M(Si-30).
n(Si-30) = 0.0091 g ÷ 29.97376 g/mol.
n(Si-30) = 0.000303 mol; amount of the silicon-30.
N(Si-30) = n(Si-30) · Na (Avogadro constant).
N(Si-30) = 0.000303 mol · 6.022·10²³ 1/mol.
N(Si-30) = 1.824·10²⁰.