How many molecules of CCl4 are present in 6.10 × 10−3 g of CCl4?

Molar mass of CCl₄ = (12.0 + 35.5×4) g/mol = 154.0 g/mol
Moles of CCl₄ = (6.10 × 10⁻³ g) / (154.0 g/mol) = 3.961 × 10⁻⁵ mol

Avogadro constant = 6.02 × 10²³ /mol
No. of CCl₄ molecules = (3.961 × 10⁻⁵ mol) × (6.022 × 10²³ molecules/mol) = 2.39 × 10¹⁹ molecules

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OR:

(6.10 × 10⁻³ g CCl₄) × (1 mol CCl₄ / 154.0 g CCl₄) × (6.022 × 10²³ CCl₄ molecules / 1 mol CCl₄)
= 2.39 × 10¹⁹ CCl₄ molecules