  The congruence subgroup problem for a finitely generated group $\Gamma$ and
$G\leq Aut(\Gamma)$ asks whether the map $\hat{G}\to Aut(\hat{\Gamma})$ is
injective, or more generally, what is its kernel $C\left(G,\Gamma\right)$? Here
$\hat{X}$ denotes the profinite completion of $X$. In this paper we investigate
$C\left(IA(\Phi_{n}),\Phi_{n}\right)$, where $\Phi_{n}$ is a free metabelian
group on $n\geq4$ generators, and $IA(\Phi_{n})=\ker(Aut(\Phi_{n})\to
GL_{n}(\mathbb{Z}))$.
  We show that in this case $C(IA(\Phi_{n}),\Phi_{n})$ is abelian, but not
trivial, and not even finitely generated. This behavior is very different from
what happens for free metabelian group on $n=2,3$ generators, or for finitely
generated nilpotent groups.
