Consider the following LTI system with impulse response $h[n]$
$$ \{v[n]\} \longrightarrow \boxed{H(z)} \longrightarrow \{u[n]\} $$
From the analysis of LTI systems with WSS random inputs, the following results deduced. Given the input $v[n]$ which is zero mean white WSS random process, its ACS (Auto-Correlation Sequence) sequence is:
$$\phi_{vv}[m] = \sigma_v^2 \delta[m]$$ where $\sigma_v^2$ variance of the input.
Then the output ACS is given by:
$$\phi_{uu}[m] = h[m] \star h[-m]^{*} \star \phi_{vv}[m]$$
Similary the input and output PSD (Power Spectral Densities) are related by:
$$S_{uu}(w) = H(w) H(w)^{*} S_{vv}(w) = |H(w)|^2 S_{vv}(w) $$ which follows from the relation that PSD of a WSS RP is the DTFT of its ACS; i.e.,
$$ S_{uu}(w) =\sum_{m=-\infty}^{\infty} \phi_{uu}[m] e^{-j w m} \longleftrightarrow \phi_{uu}[m] = \frac{1}{2\pi} \int_{-\pi}^{\pi} S_{uu}(w) e^{j w m} dw$$
Then the variance of the output is given by the following:
$$ \sigma_{uu}^2 = \phi_{uu}[0] = \frac{1}{2\pi} \int_{-\pi}^{\pi} S_{uu}(w)dw$$
Now using the given information $H(z) = \frac{1}{1-a z^{-1}}$ and $\phi_{vv}[m] = 1 \delta[m]$ you can compute the variance in either way:
$$ \sigma_u^2 = \left( h[m] \star h[-m]^{*} \star \delta[m] \right) |_{m=0} $$ or
$$ \sigma_u^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |\frac{1}{1-a e^{-jw}}|^2 dw $$
For example chosing the first approach with $h[n] = a^n u[n]$ yields:
$$
\begin{align}
\sigma_u^2 &= \left( h[m] \star h[-m]^{*} \star \delta[m] \right) |_{m=0} \\
\sigma_u^2 &= \left( h[m] \star h[-m]^{*} \right) |_{m=0} \\
\sigma_u^2 &= \sum_{k=-\infty}^{\infty} h[m-k]h[-k]^{*} |_{m=0} \\
\sigma_u^2 &= \sum_{k=-\infty}^{\infty} h[-k]h[-k]^{*} \\
\sigma_u^2 &= \sum_{k=-\infty}^{\infty} a^{-k} u[-k] (a^{-k})^{*} u[-k]\\
\end{align}
$$
$$ \sigma_u^2 = \sum_{k=-\infty}^{0} (a a^{*})^{-k} = \sum_{k=-\infty}^{0} |a|^{-2k} = \sum_{k=0}^{\infty} (|a|^{2})^k = \frac{1}{1-|a|^2}$$
