If the length of a rectangle is $$\left(\frac{2}{3}a^2 + \frac{2}{9}\right)$$ m and its breadth is $$\left(a^2 + \frac{1}{6}\right)$$ m, then the area of the rectangle (in $$\text{m}^2$$) is

A

$$\frac{2}{3}a^4 + \frac{1}{3}a^2 + \frac{1}{27}$$

B

$$\frac{2}{3}a^4 + \frac{2}{9}a^2 + \frac{1}{27}$$

C

$$\frac{2}{3}a^4 + \frac{1}{3}a^2 + \frac{2}{27}$$

D

$$\frac{2}{3}a^4 + \frac{1}{18}a^2 + \frac{1}{27}$$