= Solution
For the curvature two-form
$$
F=dA+A\wedge A,
$$
use the normalization
$$
\boxed{C_2=\frac1{8\pi^2}\operatorname{Tr}(F\wedge F)}
$$
for the <Second Chern form>. Graded cyclicity of the trace gives $\operatorname{Tr}(A^{\wedge4})=0$ and
$$
\operatorname{Tr}(F\wedge F)
=\operatorname{Tr}\left(dA\wedge dA
+2dA\wedge A\wedge A\right).
$$
On the other hand,
$$
d\operatorname{Tr}(A\wedge dA)=\operatorname{Tr}(dA\wedge dA),
$$
and
$$
d\operatorname{Tr}(A\wedge A\wedge A)
=3\operatorname{Tr}(dA\wedge A\wedge A).
$$
Thus the coefficient in the <Chern-Simons 3-form> must be
$$
\boxed{c=\frac23},
$$
and
$$
\boxed{C_2=dY,\qquad
Y=\frac1{8\pi^2}\operatorname{Tr}\left(
A\wedge dA+\frac23A\wedge A\wedge A\right)}.
$$
For the gauge transformation $A'=gAg^{-1}-dg\,g^{-1}$, use $d(g^{-1})=-g^{-1}(dg)g^{-1}$ and the <Maurer-Cartan equation>. Expanding $dA'+A'\wedge A'$ makes the terms linear and quadratic in $dg\,g^{-1}$ cancel, leaving
$$
\boxed{F'=gFg^{-1}}.
$$
Invariance of the matrix trace under conjugation then gives
$$
\boxed{C_2'=\frac1{8\pi^2}\operatorname{Tr}(gFg^{-1}\wedge gFg^{-1})=C_2}.
$$
A <Yang-Mills instanton> on $\mathbb R^4$ has finite Euclidean action, smooth curvature in the interior, and $F\to0$ sufficiently rapidly at infinity. Its connection therefore approaches a pure gauge on the asymptotic three-sphere,
$$
A\longrightarrow-dg\,g^{-1}
$$
up to a decaying correction, for a map $g:S^3_\infty\to SU(2)$. By <Stokes theorem>,
$$
k=\int_{\mathbb R^4}C_2
=\int_{S^3_\infty}Y.
$$
Writing $\theta=dg\,g^{-1}$ gives $d\theta=\theta\wedge\theta$ and, for $A=-\theta$,
$$
Y=\frac1{24\pi^2}\operatorname{Tr}(\theta^{\wedge3}).
$$
Hence the <instanton number as a winding number at infinity> is
$$
\boxed{k=\frac1{24\pi^2}\int_{S^3_\infty}
\operatorname{Tr}\left[(dg\,g^{-1})^{\wedge3}\right]\in\mathbb Z}.
$$
Under $SU(2)\cong S^3$, this integer is the <degree of a map between oriented manifolds> $g:S^3\to S^3$. Reversing the trace or orientation convention reverses the displayed sign.
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