Mole concept -2

$\begin{aligned} \text{300 ppm of }\ce{CaCO3} \text{ in water} & = \frac{\text{300 g }\ce{CaCO3}}{10^6 \text{ g }\ce{H2O}} \\ & = \frac{\text{0.3 g }\ce{CaCO3}}{1000 \text{ g }\ce{H2O}} \\ & \approx \frac{\text{0.3 g }\ce{CaCO3}}{1 \text{ L }\ce{H2O}} \\ & \approx \frac{\color{#3D99F6}{\text{0.003 mol }}\ce{CaCO3}}{1 \text{ L }\ce{H2O}} \quad \quad \small \color{#3D99F6}{\text{Since molar mass of }\ce{CaCO3} \approx \text{100 g}} \\ & = \boxed{0.003} \text{ M} \end{aligned}$

300 ppm Means 300 mg per Litre. That means 0.3 gm CaCO3 or (0.3/100)=0.003 mole CaCO3 per Litre. And that is the required molarity 0.003.