The correct answer is $\boxed{(0.2)u[-n-1]}$.
The z-transform of the impulse response of a system is the Laplace transform of the system’s impulse response evaluated at $s=j\omega$, where $\omega$ is the angular frequency. The ROC of the z-transform is the set of all $z$ values for which the z-transform converges.
In this case, the z-transform is given by
$$H(z) = \frac{z}{z-0.2}$$
The ROC is $|z| < 0.2$. This means that the z-transform converges for all $z$ values such that $|z| < 0.2$.
The inverse z-transform of $H(z)$ is given by
$$h[n] = \frac{1}{2\pi i} \oint_C \frac{z^{n-1}}{z-0.2} dz$$
where $C$ is a circle of radius $0.2$ centered at the origin.
Evaluating the integral, we get
$$h[n] = (0.2)u[-n-1]$$
where $u[n]$ is the unit step function.
Therefore, the impulse response of the system is $(0.2)u[-n-1]$.
Option A is incorrect because it does not satisfy the ROC. Option B is incorrect because it is the negative of the impulse response. Option C is incorrect because it is the negative of the impulse response with a delay of 1 sample. Option D is incorrect because it is the impulse response with a delay of 1 sample.