If x and y are natural numbers such that $$(\frac{1}{x})^\frac{1}{y}=0.\overline{3}$$, then the value of XY is

$$0.\overline{3}$$ can be written as $$\frac{1}{3}$$. So the possible values of x, y will be as follows.

3,1 $$\Rightarrow\ (\frac{1}{x})^{\frac{1}{y}}=\frac{1}{3}$$. Here, xy=3.

9,2 $$\Rightarrow\ (\frac{1}{x})^{\frac{1}{y}}=\left(\frac{1}{9}\right)^{\frac{1}{2}}=\frac{1}{3}$$. Here, xy=18.

27,3$$\Rightarrow\ (\frac{1}{x})^{\frac{1}{y}}=\left(\frac{1}{27}\right)^{\frac{1}{3}}=\frac{1}{3}$$. Here, xy=81.

So, option B is correct.