Estimating the white noise variance This is a mock-exam question:

The following measurements have been made
$\hat{\gamma}(0) = 570, \quad \hat{\gamma}(1) = 505, \quad \hat{\gamma}(2) = 460, \quad \hat{\gamma}(3) = 420$
Suppose that the data was generated from an $AR(2)$ model,
$X_t - \phi_1 X_{t-1} - \phi_2 X_{t-2} = Z_t$
where $\{Z_t\} \sim WN(0,\sigma^2)$. Estimate the variance $\sigma^2$.

What i've tried is simply to use the Yule-Walker equations to estimate the variance,
\begin{equation*}
\gamma(k) - \phi_1 \gamma(k - 1) - ... - \phi_p \gamma(k - p) = 
\begin{cases}
0,        \quad &\text{if } k = 1, ..., p \\
\sigma^2, \quad &\text{if } k = 0
\end{cases}
\end{equation*}
which, if $k = 0$, should give $\gamma(0) = \hat \gamma(0) = \sigma^2 = 570$. But the answer is $\sigma^2 = 121$.
What am I missing here?

$\gamma(0)$ is the variance of $X_t$, it is not in general equal to the variance of $Z_t$, which is what you're looking for.
You may also be misinterpreting the above equation. For $p=2$ and $k=0$, it says that:
$$\gamma(0) - \phi_1 \gamma(-1) - \phi_2 \gamma(-2) = \sigma^2$$
It does not say that $\gamma(0) = \sigma^2$. The terms where $\gamma(i)$ has a negative argument are not implicitly dropped or anything like that. Since $\gamma$ is symmetric, it also says that:
$$\gamma(0) - \phi_1 \gamma(1) - \phi_2 \gamma(2) = \sigma^2$$
You can combine this with the same equation for $k=1$ and $k=2$ to build a linear system with 3 equations and 3 unknowns, something like this:
$$\begin{pmatrix}
\gamma(1) && \gamma(2) && 1 \\
\gamma(0) && \gamma(1) && 0 \\
\gamma(1) && \gamma(0) && 0
\end{pmatrix}
\begin{pmatrix}
\phi_1 \\
\phi_2 \\
\sigma^2
\end{pmatrix}
=
\begin{pmatrix}
\gamma(0) \\
\gamma(1) \\
\gamma(2)
\end{pmatrix}
$$
More typically this is done in two steps, solving for the $\phi_i$ first in the smaller system:
$$\begin{pmatrix}
\gamma(0) && \gamma(1) \\
\gamma(1) && \gamma(0) \\
\end{pmatrix}
\begin{pmatrix}
\phi_1 \\
\phi_2
\end{pmatrix}
=
\begin{pmatrix}
\gamma(1) \\
\gamma(2)
\end{pmatrix}
$$
And then plugging in $\phi_i$ into the remaining equation to get $\sigma^2$.
Either way, the result is:
$$\sigma^2 = \left(\gamma(0) - \gamma(2)\right) \left( 1 + \frac{\gamma^2(1)-\gamma(0)\gamma(2)}{\gamma^2(1)-\gamma^2(0)}\right)$$
Substituting for the $\hat{\gamma}(i)$, you'll get to $\hat{\sigma}^2 \approx 121$.
