10L of hard water required 0.56g of lime for removing hardness. Hence temporary hardness in ppm
of CaCO₃ is

10L = 10000ml = 10⁴ g contains ... 0.56g 
10⁶g ......... ? = 56g 
100g of CaCO₃ = 56g of CaO

$\begin{array}{r}0.56g\mathrm{CaO}\equiv 2g{\mathrm{CaCO}}_3\mathrm{in} 10{\mathrm{LH}}_2\mathrm{O}\\ =2g{\mathrm{CaCO}}_3\mathrm{in} {10}^4{\mathrm{mLH}}_2\mathrm{O}\\ =200g{\mathrm{CaCO}}_3\mathrm{in} {10}^6{\mathrm{mLH}}_2\mathrm{O}\end{array}$