The number of iron (Fe) atoms present in a pure sample of solid iron with a mass of 10 grams is equal to- (The atomic mass of iron is 55.9) .

The correct option is C $\frac{\left(10.0\right)(6.02\times {10}^{23})}{\left(55.9\right)}$
The number of iron (Fe) atoms present in a pure sample of solid iron with a mass of 10 grams is equal to $\frac{\left(10.0\right)(6.02\times {10}^{23})}{\left(55.9\right)}$
Hence, the option (C) is the correct answer.
Note : $\frac{\left(10.0\right)}{\left(55.9\right)}$ represents the ratio of mass of iron to molar mass. It represents number of moles of iron.
When number of moles of iron is multiplied with avogadro's number $(6.02\times {10}^{23})$, we get number of iron atoms. (The atomic mass of iron is $55.9$) .