equation

SImplify as follows:

 $\sqrt{{3^2 \over 5^2}^{-3}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}$

${\sqrt {1 \over{3^2 \over 5^2}^3}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}$

${\sqrt {1 \over{3^6 \over 5^6}}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}$

${\sqrt {5^6 \over 3^6}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}$

${5^3 \over 3^3} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}$

$5^3 - \sqrt{x} = 11^2$

$125 - \sqrt{x} = 121$

$\sqrt{x} = 4$

$\color{brown}\boxed{x = 16}$

Thanks Builderboi,   

Here ia another approach. 

$\left(\frac{9}{25}\right)^{-\frac{3}{2}}-\frac{\sqrt x}{27}=\left(\frac{3^3}{11^2}\right )^{-1}\\~\\ \left(\frac{3}{5}\right)^{-3}-\frac{\sqrt x}{27}=\left(\frac{27}{121}\right )^{-1}\\~\\ \left(\frac{5}{3}\right)^{3}-\frac{\sqrt x}{27}=\frac{121}{27}\\~\\ \frac{125}{27}-\frac{\sqrt x}{27}=\frac{121}{27}\\~\\ 125-\sqrt x = 121\\~\\ \sqrt x = 4\\~\\ x=16$