what's the mass of the original sample (g)?
It is first of all important to recognise that all the chloride in the original sample will precipitate upon treatment with excess AgNO3. This will give you a starting point, which is a measurable amount of AgCl.

The molecular mass of AgCl = (107.87+35.45)=143.32g

Therefore, the proportion of chloride in AgCl = (35.45/143.32) = 0.2473

Therefore in 2.013g of AgCl, there is (0.2473 x 2.013) = 0.4978g of chloride.

If 0.4978g of chloride constitutes 75% of the mass of the original sample, then the amount of original sample is (0.4978 / 0.75) = 0.6637g