What value comes in place of question mark (?) in the following statements and arrange them in descending order.
$$A. \dfrac{5}{9}\times(225.4-45.4)=(?)^{2}$$
$$B.\sqrt{1024}\times40+(20)^2+0.5\%$$ of $$9600+469=(?)^{3}$$
$$C. 8^{12} \div 16^{2}$$ of $$32^{3}\times\sqrt{256}=(2)^{?}$$
$$D. 18{\frac{1}{3}} + 9 {\frac{2}{3}} - 10 {\frac{1}{3}}=1 {\frac{2}{3}}+?$$
Choose the correct answer from the options given below :

Evaluating A:

$$\dfrac{5}{9}\times\ \left(225.4-45.4\right)=\dfrac{5}{9}\times\ 180=100=\left(10\right)^2$$

So, A= 10

Evaluating B:

$$\sqrt{1024}\times40+(20)^2+0.5\%$$ of $$9600+469=(?)^{3}$$

=$$32\ \times\ 40+400+\dfrac{0.5}{100}\times\ 9600+469=32\times\ 40+400+48+469=2197=13^3$$

So, B=13

Evaluating C:

$$\dfrac{8^2}{16^2}\times\ 32^3\times\ \sqrt{\ 256}=\dfrac{1}{2^2}\times\ \left(2^5\right)^3\times\ 2^4=2^{17}$$

So, C=17

Evaluating D:

$$18\dfrac{1}{3}+9\dfrac{2}{3}-10\dfrac{1}{3}=8+9\dfrac{2}{3}=1\dfrac{2}{3}+8+8=1\dfrac{2}{3}+16$$

So, D = 16

So, the correct order of arrangement is $$C>D>B>A$$