= Solution
Average the trial field over an integer number of periods. The squared-field, squared-gradient and squared-Laplacian averages are $A^2/2$, $q_0^2A^2/2$ and $q_0^4A^2/2$, while $\langle\phi^2|\nabla\phi|^2\rangle=A^4q_0^2/8$. Hence the <single-mode smectic mean-field free energy> is
$$
\boxed{\frac HV=\frac{a-a_c}{4}A^2+\frac{Bq_0^2}{8}A^4.}
$$
For the physically stabilizing case $B>0$, minimization gives
$$
\boxed{A^2=\begin{cases}0,&a\geq a_c,\\(a_c-a)/(Bq_0^2),&a<a_c,\end{cases}\qquad a_c=\frac{\kappa^2}{4\gamma}.}
$$
The amplitude tends continuously to zero at $a_c$, so this <mean-field approximation> predicts a <continuous phase transition>. The ordered minimum has <free-energy density> $-(a_c-a)^2/(8Bq_0^2)$.
A nonzero $B$ alone is not sufficient for this conclusion: $B<0$ makes the displayed quartic amplitude energy unbounded below and requires higher-order stabilization. The continuous-transition statement uses $B>0$ and the stipulated fixed wavevector $q_0$.
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