d
\(16.0\,g\,{O_3} = \frac{{16}}{{48}}\) mole

\( = \frac{{16}}{{48}} \times 6.023 \times {10^{23}}\) molecules

\( = 3 \times \frac{{16}}{{48}} \times 6.023 \times {10^{23}}\) atoms

\(= 6.023 \times 10^{23}\) atoms

\(28.0\,g\,CO = \frac{{28}}{{28}}\) mole \(= 1\) mole

\( = 1 \times 6.023 \times {10^{23}}\) molecules

\( = 1 \times 6.023 \times {10^{23}}\) atoms

\( = 6.023 \times {10^{23}}\) atoms

\(16.0\,g\,{O_2} = \frac{{16}}{{16}}\) mole \(= 1\) mole

\( = 1 \times 6.023 \times {10^{23}}\) molecules

\( = 1 \times 6.023 \times {10^{23}}\) atoms

\( = 6.023 \times {10^{23}}\) atoms

Therefore, the ratio is

\(6.023 \times {10^{23}}:6.023 \times {10^{23}}:6.023 \times {10^{23}}\)

i.e. \(1:1:1\)