$Ag_2CO_3 \underrightarrow{\Delta} Ag_2O$

In that $Ag_2CO_3$ is $ 2.76\;g ; \qquad Ag_2O$ is $xg$

Molecular mass of $Ag_2CO_3$

$2(108) +(12)+3(16)=276$

Number of moles of $Ag_2CO_3=\large\frac{2.76}{276}=$$0.01 \;mol$

Now 1 mole of $Ag_2CO_3$ releases 1 mole of $w_2$ and 1 mole of $Ag_2O$ remains as residue.

$\therefore $ residue $=0.01 \times $molecular mass of $Ag_2O$

$\qquad= 0.01 \times 232=2.32\;g$

Hence c is the correct answer.