No. of moles of $A=\frac{6.0\,g}{60\,g/mol}=0.1\,mol$

No. of moles of $B=\frac{6.00\times {{10}^{23}}}{6.000\times {{10}^{23}}}=1\,mol$

No. of moles of $C = 0.036$

$AB_2C_3$ formed accordingly to $C$ which is a limiting reagent.

Since $3$ moles of $C$ are used in $(1)$

So it gives $1$ mole of $AB_2C_3$

$n_{A{{B}_{2}}{{C}_{3}}}=\frac{0.036}{3}=0.012$

$=\frac{Given\,mass\,(4.8)}{Molecular\,mass\,(M.M)}$

Mol.mass $=\frac{4.8}{0.012}=400$

$\Rightarrow 400=60+(2\times x)+(80\times 3)$

$\Rightarrow x=50$