${\mathrm{Na}}_2{\mathrm{CO}}_3$ does not decompose on heating.

${\mathrm{CaCO}}_3$ decomposes to form $\mathrm{CaO}$ and ${\mathrm{CO}}_2$.

${\mathrm{CaCO}}_3\stackrel{\Delta }{⟶}\mathrm{CaO}+{\mathrm{CO}}_2$

$\therefore$ Loss in mass is due to loss of ${\mathrm{CO}}_2$ gas

$\therefore (3.88-3.00=0.88\mathrm{g})$ of ${\mathrm{CO}}_2$ is lost during heating.

$0.88 g {\mathrm{CO}}_2=\frac{0.88}{44}=0.02 \mathrm{mol} {\mathrm{CO}}_2$

From the reaction stoichiometry,

$0.02 \mathrm{mol} {\mathrm{CO}}_2\equiv 0.02 \mathrm{mol} {\mathrm{CaCO}}_3$

$0.02 \mathrm{mol} {\mathrm{CaCO}}_3=0.02\times 100=2.00g {\mathrm{CaCO}}_3$

$\therefore$ Mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ in the mixture $=3.88-2.00 \mathrm{g}=1.88 \mathrm{g}$