Incorrect. Factor the radicand by separating out all the perfect squares:

$\displaystyle\sqrt{{{108}{{x}}^{{11}}{{y}}^{{4}}{{z}}^{{9}}}}=\sqrt{{{36}\cdot{3}\cdot{{x}}^{{10}}\cdot{x}\cdot{{y}}^{{4}}\cdot{{z}}^{{8}}\cdot{z}}}\usepackage{cmbright}\Huge\$

Rewrite all squares as a number to the power 2; for example, $\displaystyle{{x}}^{{10}}={{\left({{x}}^{{5}}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ .

Separate into individual radicals using $\displaystyle\sqrt{{{a}{b}}}=\sqrt{{a}}\cdot\sqrt{{b}}\usepackage{cmbright}\Huge\$ . Simplify and use absolute values where necessary. Multiply to get $\displaystyle{6}{\left|{{x}}^{{5}}\right|}{{y}}^{{2}}{{z}}^{{4}}\sqrt{{{3}{x}{z}}}\usepackage{cmbright}\Huge\$ . The correct answer is $\displaystyle{3}{x}{z}\usepackage{cmbright}\Huge\$ .