Which of the following is correct?

Correct option is A

Given: 
$$\sqrt[6]{8},\quad \sqrt[4]{3},\quad \sqrt[3]{2},\quad \sqrt[12]{12}$$ 
Solution:
​$$\sqrt[6]{8},\quad \sqrt[4]{3},\quad \sqrt[3]{2},\quad \sqrt[12]{12}\\= 8^{\frac{1}{6}},\quad 3^{\frac{1}{4}},\quad 2^{\frac{1}{3}},\quad 12^{\frac{1}{12}}\\\text{(LCM of denominators = 12)}\\\text{Multiply all exponents by 12:}\\(8^{\frac{1}{6}})^{12} = 8^2 = 64, \\ (3^{\frac{1}{4}})^{12} = 3^3 = 27\\ (2^{\frac{1}{3}})^{12} = 2^4 = 16\\ (12^{\frac{1}{12}})^{12} = 12\\\text{So, } 64 > 27 > 16 > 12\\\Rightarrow \sqrt[6]{8} > \sqrt[4]{3} > \sqrt[3]{2} > \sqrt[12]{12}$$​​