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Explanation:

Given, ${\log }_{\sqrt{3}} \left(x\right)$ + $\frac{{\log }_x 25}{{\log }_x (0.008)}$ = $\frac{16}{3}$

⇒ ${\log }_{3^{1/2}} \left(x\right)$ + $\frac{{\log }_x 5^2}{{\log }_x {\left(5\right)}^{-3}}$ = $\frac{16}{3}$

⇒ $2\times {\log }_3 \left(x\right)$ + $\frac{2\times {\log }_x 5}{-3\times {\log }_x \left(5\right)}$ = $\frac{16}{3}$

⇒ $2\times {\log }_3 \left(x\right)$ - $\frac{2}{3}$ = $\frac{16}{3}$

⇒ $2\times {\log }_3 \left(x\right)$ = $\frac{16}{3}$ + $\frac{2}{3}$ = 6

⇒ log₃ x = 3

⇒ x = 27

Now, ${\log }_3 \left(3x^2\right)$ = ${\log }_3 (3\times {27}^2)$ = ${\log }_3 \left(3^7\right)$ = 7

Hence, 7.