= Solution
{c}
For fixed $A$, the delta functional sets $\sigma(x)=F[A](x)$. Therefore its weighted <functional integral> is
$$
\int\mathcal D\sigma\,\delta(F[A]-\sigma)\exp\left[-\frac{i}{2\xi}\int d^4x\,\sigma^2\right]
=\exp\left[-\frac{i}{2\xi}\int d^4x\,F[A]^2\right].
$$
Since the action enters as $e^{iS}$, the <gauge fixing> contribution is
$$
\boxed{\mathcal L_{\mathrm{gf}}=-\frac1{2\xi}\left(\partial_\mu A^\mu+\lambda A_\mu A^\mu\right)^2.}
$$
The <Faddeev-Popov determinant> does not depend on $\sigma$ and remains in the <path integral>. The limit $\xi\to0$ enforces the original condition sharply; for nonzero $\xi$ this is a weighted average of the gauge slices.
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