The weight of $MgC{O}_3$ required for the preparation of $12g$ of $MgS{O}_4$ by reacting with sulphuric acid is

 The balanced equation is 

$MgC{O}_3+H_{2}S{O}_4\to MgS{O}_4+C{O}_2+H_2{O}_{ }$

From the equation we can conclude that 

$1$ mole of $MgC{O}_3$ produces $1$ mole of $MgS{O}_4$

 $1$ mole $MgC{O}_3$ weight is $84g$

$1$ mole $MgS{O}_4$ weight is $120g$       

$84g\to 120g$

$xg\leftarrow 12g$

Upon cross multiplication, we get 

$x\times 120=12\times 84$

$x=12/120\times 84$

$x=0.1\times 84$

$x=8.4$

Hence the correct option is (A).