Correct option is A. 200
According to the given reaction
$$ RH_{2}+Ca^{2+}\rightarrow RCa+2H^{+} $$
Each mole $$ Ca^{2+} $$ ion replaced by 2 moles $$ H^{+}$$
1 mole $$ H^{+} $$ replaced $$ \Rightarrow \dfrac{1}{2} = 0.5\,mole \,Ca^{2+} $$
Given,
$$ pH = 2 $$
$$ H^{+} = 10^{-2} = 0.01 $$
0.01 mole $$ H^{+} $$ replaced $$ = 0.01\times 05 = 0.005\,moles\,Ca^{2+} $$
Mass $$ Ca^{2+}$$ replaced $$ = 0.005\times 40 = 0.2\,g = 200\,mg $$
Concentration or Hardness of $$ Ca^{2+} = 200\,mg/L $$
$$ = 200\,ppm $$
Hence, the correct option is $$\text{A}$$