Rewrite using fractional exponents, then apply the laws of exponents:

$\displaystyle\frac{{{{s}}^{{3}}{{\left({36}{{r}}^{{9}}{{s}}^{{4}}\right)}}^{{\frac{{1}}{{2}}}}}}{{{4}{s}{{\left({27}{{r}}^{{12}}\right)}}^{{\frac{{1}}{{3}}}}}}=\frac{{{{s}}^{{3}}\cdot{{\left({{6}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\cdot{{r}}^{{\frac{{9}}{{2}}}}\cdot{{s}}^{{\frac{{4}}{{2}}}}}}{{{4}\cdot{s}\cdot{{\left({{3}}^{{3}}\right)}}^{{\frac{{1}}{{3}}}}\cdot{{r}}^{{\frac{{12}}{{3}}}}}}=\frac{{{{s}}^{{3}}\cdot{6}\cdot{{r}}^{{\frac{{9}}{{2}}}}\cdot{{s}}^{{2}}}}{{{4}\cdot{s}\cdot{3}\cdot{{r}}^{{4}}}}=\frac{{{6}\cdot{{s}}^{{5}}\cdot{{r}}^{{\frac{{9}}{{2}}}}}}{{{12}\cdot{s}\cdot{{r}}^{{4}}}}\usepackage{cmbright}\Huge\$

Rewrite as a product: $\displaystyle\frac{{6}}{{12}}\cdot\frac{{{s}}^{{5}}}{{s}}\cdot\frac{{{r}}^{{\frac{{9}}{{2}}}}}{{{r}}^{{4}}}=\frac{{1}}{{2}}\cdot{{s}}^{{{5}-{1}}}\cdot{{r}}^{{\frac{{9}}{{2}}-{4}}}=\frac{{{{s}}^{{4}}{{r}}^{{\frac{{1}}{{2}}}}}}{{2}}\usepackage{cmbright}\Huge\$

Change the $\displaystyle{{r}}^{{\frac{{1}}{{2}}}}\usepackage{cmbright}\Huge\$ to radical form: $\displaystyle\frac{{{{s}}^{{4}}\sqrt{{r}}}}{{2}}\usepackage{cmbright}\Huge\$