= Solution
For a laterally infinite slab in a parallel field there is no demagnetizing correction. The field has only its applied direction, varies with $x$, and satisfies $B''=B/\lambda^2$. Reflection <symmetry> gives $B'(0)=0$, and continuity gives $B(\pm a)=B_a$. Thus
$$
\boxed{\frac{B(x)}{B_a}=\frac{\cosh(x/\lambda)}{\cosh(a/\lambda)}}.
$$
Uniformly for $|x|\le a\ll\lambda$, expanding numerator and denominator to second order yields
$$
\boxed{\frac{B(x)}{B_a}=\frac{2\lambda^2+x^2}{2\lambda^2+a^2}+O((a/\lambda)^4)}.
$$
Only a small fraction of the applied field is screened in this limit; the almost uniform internal field is the origin of the enhanced parallel critical field.
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