If ${\log }_{\mathrm{a}}\mathrm{b}=2;{\log }_{\mathrm{b}}\mathrm{c}=2\text{ and }{\log }_3\mathrm{c}=3+{\log }_3\mathrm{a}$ then the value of $\frac{\mathrm{c}}{\left(\mathrm{ab}\right)}$ is

$\begin{array}{l}{\log }_3\mathrm{c}=3+{\log }_3\mathrm{a}\Rightarrow {\log }_3\frac{\mathrm{c}}{\mathrm{a}}=3\Rightarrow \mathrm{c}=27\mathrm{a}---\left(\mathrm{i}\right)\\ {\log }_{\mathrm{a}}\mathrm{b}=2;{\log }_{\mathrm{b}}\mathrm{c}=2\\ \Rightarrow {\log }_{\mathrm{a}}\mathrm{b}\cdot {\log }_{\mathrm{b}}\mathrm{c}=4\Rightarrow {\log }_{\mathrm{a}}\mathrm{c}=4\Rightarrow \mathrm{c}={\mathrm{a}}^4---\left(\mathrm{ii}\right)\end{array}$

From Eqs. (i) and (ii), we get a= 3, c = 81.
From relation (i), we have b = a² = 9.
Hence, $\frac{\mathrm{c}}{\left(\mathrm{ab}\right)}$ = 3.