Amanita, let's solve the problem step by step. The problem asks us to solve the inequality:

$\left(\frac{1}{3}\right)^x \geq \left(\frac{1}{3}\right)^{0.2}$

and choose the correct interval for $x$.

Step-by-Step Plan

1. Understand the inequality: Both sides have the same base, $\frac{1}{3}$, which is less than 1. For exponential inequalities with a base less than 1, the direction of the inequality changes when comparing exponents.
2. Compare the exponents: Since the bases are the same and less than 1, the inequality $\left(\frac{1}{3}\right)^x \geq \left(\frac{1}{3}\right)^{0.2}$ is equivalent to $x \leq 0.2$.
3. Write the solution as an interval: Express the solution in interval notation.
4. Check the answer choices: Match the correct interval from the provided options.

Let's solve the inequality for $x$ and find the correct interval. Amanita, here is the detailed solution:

1. We start with the inequality:

$\left(\frac{1}{3}\right)^x \geq \left(\frac{1}{3}\right)^{0.2}$

2. Since the base $\frac{1}{3}$ is less than 1, the inequality sign reverses when comparing the exponents. This means:

   $$x \leq 0.2$$

3. In interval notation, this solution is:

   $$x \in (-\infty, 0.2]$$

4. Looking at the answer choices, the correct interval is:

   $$x \in (-\infty; 0.2]$$

So, the answer is: $x \in (-\infty; 0.2]$.