One mole of CaCO3 reacts with two moles of HCl. You first find the number of moles of CaCO3 that can react with the HCl. Since 6.3g gives 0.063 moles of CaCO3 you then need to find the amount of HCl that will react.
6.3g of CaCO3/100g of CaCO3=0.063
1 mole of CaCO3 will react with 2 moles of HCl : 0.063 of CaCO3 * 2= 0.126 moles of HCl
You then find what volume of HCl will react.
Vol =Mol/Mol con
Vol=0.126moles/0.55M=0.229L .It is in L since we are using M so you convert to mL .
0.229L*1000=229mL of HCl