= Solution
Use the <wave-function renormalization> coefficient and <anomalous dimension> definition stipulated in this part, keeping their signs and normalization. Put $A=1/[6(32\pi^2)^2]$. To the required order,
$$
\log Z=A\frac{g_4^2}{\epsilon}+O(g_4^3).
$$
At fixed bare $\lambda_4$, the leading change of the dimensionless coupling is $\mu\,dg_4/d\mu=-2\epsilon g_4+O(g_4^2)$, because $\lambda_4$ has <mass dimension> $2\epsilon$. Therefore the chain rule, applied before setting $\epsilon=0$, gives
$$
\begin{aligned}
\gamma(g_4)&=-\beta_\epsilon(g_4)\frac{\partial\log Z}{\partial g_4}\\
&=-(-2\epsilon g_4)\frac{2Ag_4}{\epsilon}+O(g_4^3)
=4Ag_4^2+O(g_4^3).
\end{aligned}
$$
Thus
$$
\boxed{\gamma(g_4)=\frac{g_4^2}{6(16\pi^2)^2}+O(g_4^3).}
$$
The vanishing engineering contribution $-2\epsilon g_4$ multiplies a pole and leaves a finite answer. Dropping it at the start would incorrectly give zero. The numerical answer is conditional on the supplied two-loop coefficient. A convention defining the field anomalous dimension as $\gamma_\phi=\tfrac12\mu\,d\log Z/d\mu$ instead gives $\gamma_\phi=-\gamma/2$ for the same $Z$; substituting that convention silently would answer a different question.
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