Judging by the formula M2CO3, the ion charge on M is a +1.

1 M2CO3 + 2 HCl ==> 1 MCl2 + 1 CO2 + 1 H2O

560 cm3 CO2 ==> 560/22400 mol of CO2 (because 1 mol of any gas occupies 22400 cm3 at standard temperature and pressure - note, standard and not room)

560 cm3 CO2 ==> 0.025 mol CO2

Using the mole ratio 1 : 1. it is quite obvious that 0.025 mol of CO2 must be produced from 0.025 mol of M2CO3.

Equivalent mass of M2CO3 in 0.025 mol M2CO3 is 2.65 g.

So, relative mass of M2CO3
= 2.65 / 0.025
= 106

Subtract away 12 (for the carbon atom) and 3 x 16 (for the three oxygen atoms), we get 46 for the two M atoms.

[M is sodium if you work it out]

So, percentage by mass of metal M
= 46/106 x 100%
= 43.396...%
~ 43.4%