= Solution
Average the exact slab solution across its thickness:
$$
\overline B=B_a\frac\lambda a\tanh(a/\lambda),\qquad M=\frac{\overline B-B_a}{\mu_0}=-\frac{B_a}{\mu_0}\left[1-\frac{\tanh u}{u}\right],\quad u=a/\lambda.
$$
For $u\ll1$, $M\simeq-B_a a^2/(3\mu_0\lambda^2)$. The zero-field condensation <free energy> per volume is $-B_C^2/(2\mu_0)$. Neglecting normal-state magnetic susceptibility, the Gibbs free-energy difference at fixed applied field follows from $d(G_s-G_n)/dB_a=-M$:
$$
G_s-G_n=-\frac{B_C^2}{2\mu_0}+\frac{B_a^2}{2\mu_0}\left(1-\frac{\tanh u}{u}\right).
$$
Equating it to zero gives the constant-$\lambda$ estimate for the <parallel critical field of a thin London slab>
$$
\boxed{\frac{B_{\parallel C}}{B_C}=\left(1-\frac{\tanh(a/\lambda)}{a/\lambda}\right)^{-1/2}\simeq\frac{\sqrt3\lambda}{a}}.
$$
The enhancement is equivalently $\sqrt{12}\lambda/d$ for total thickness $d=2a$. It is an estimate assuming an intact, spatially uniform condensate density up to the transition. Field-induced pair suppression, the paramagnetic limiting field, and failure of bulk local electrodynamics in an extremely thin film limit an unlimited extrapolation as $a\to0$.
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