\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.30^{14}\right)^2}=\frac{11.3^{22+7}-\left(3^2\right)^{15}}{2^2.30^{28}}=\frac{11.3^{29}-3^{30}}{2^2.\left(3.10\right)^{28}}=\frac{11.3^{29}-3.3^{29}}{2^2.10^{28}.3^{28}}\)
\(=\frac{3^{29}\left(11-3\right)}{2^2.10^{28}.3^{28}}=\frac{3.2}{10^{28}}=\frac{6}{10^{28}}\)