Noise added to a Random Process
Hi: I don't know whether white noise can have a non-zero mean ( Dilipe can tell us that )  but, if it can't, then just call it whatever you call white noise with a non-zero mean.
But, assuming it's zero, this is how you can do it. ( If it's not zero, then the  solution is similar but the last step would be a little different. Use
$$\operatorname{Var}\left\{X\right\} = E\left\{X^2\right\} - E\left\{X\right\}^2$$
Note also I wrote it with &quot;t's&quot; and &quot;epsilon's&quot;  just because I'm more used to those but you can think of $t$ as $n$ and $\epsilon$ as $w$.
\begin{align}
x_t &amp;= 0.2 x_{t-1} + \epsilon_{t} + \epsilon_{t-1}\\
\implies x_{t}(1 - 0.2L) &amp;= \epsilon_{t} + \epsilon_{t-1} \tag{$\scriptsize{\text{using lag operator}}$}\\
\implies x_{t} &amp;=  \frac{\epsilon_{t}}{1 - 0.2L} + \frac{\epsilon_{t-1}}{1 - 0.2L}\\
\implies x_{t} &amp;= \sum_{i = 0}^{\infty} 0.2^{i} \epsilon_{t-i} + \sum_{i = 0}^{\infty} 0.2^{i} \epsilon_{t-1-i}
\end{align}
So, since the mean is zero, the $E\left\{x^2_{t}\right\}$ is the variance of the RHS.
So, it is
\begin{align}
\sum_{i = 0}^{\infty} \left(0.2^{i}\right)^2 \cdot \sigma^2_\epsilon + \sum_{i = 0}^{\infty} \left(0.2^{i}\right)^2 \cdot \sigma^2_\epsilon&amp;= \sum_{i = 0}^{\infty} 0.04^{i} \cdot \sigma^2_\epsilon + \sum_{i = 0}^{\infty} 0.04^{i} \cdot \sigma^2_\epsilon\\
 &amp;=  \frac{2 \cdot \sigma^2_{\epsilon}}{(1- 0.04)}
\end{align}