That'd be it.
It said that the solubility = 245g AgNO3 / 100 mL
245 g AgNO3 / (169.873 g/mol) = 1.442 mol AgNO3
So solubility = 1.442 mol / 100 mL = 14.42 mol/L = 14.42 M
0.1 M AgNO3, therefore, should be REALLY easy to make. This is what tells us it's something OTHER than insolubility of AgNO3