One litre of a samle of hard water contains 1 mg of $CaC{l}_2$ and 1 mg of $MgC{l}_2$.Then the total hardness in terms of parts of $CaC{O}_3$ per $10^8$ parts of water by mass is:

$1$ mole of $CaC{l}_2\equiv 1$ mole of $CaC{O}_3\equiv 1$ mole of $MgC{l}_2$

$\therefore$ $100mg$ $CaC{O}_3$ is produced by $95mg$ $MgC{l}_2$

$\therefore$ $1mg$ of $MgC{l}_2$ gives $\frac{100}{95}mg$  $CaC{O}_3=$ $1.05mg$ $CaC{O}_3$

Similarly, $1mg$ of $CaC{l}_2$ gives $\frac{100}{111}mg$ $CaC{O}_3=$ $0.90mg$ $CaC{O}_3$

$\therefore$ Total $CaC{O}_3$ per litre of water= $1.05+0.90=1.95$ $mg$

Weight of $1000ml$ of water= $10^{3}g$=$10^{6}mg$

$\therefore$ Total hardness in terms of parts of $CaC{O}_3$ per $10^6$ parts of water by weight = $\frac{1.95}{10^6}\times 10^6$

=$1.95$ $ppm$

Answer: (A) $1.954$ $ppm$