a
\(A+3B+3C\rightleftharpoons A{{B}_{2}}{{C}_{3}}.....(1)\)

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 \(=\frac{4.8}{0.012}=400\)

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

\(\Rightarrow x=50\)