Since $ \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{b}, $ \[(\mathbf{c} + \mathbf{c} \times \mathbf{a}) \cdot (\mathbf{c} + \mathbf{c} \times \mathbf{a}) = \mathbf{b} \cdot \mathbf{b}.\]This expands as \[\mathbf{c} \cdot \mathbf{c} + 2 \mathbf{c} \cdot (\mathbf{c} \times \mathbf{a}) + (\mathbf{c} \times \mathbf{a}) \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{b} \cdot \mathbf{b}.\]We know $ \mathbf{b} \cdot \mathbf{b} = \|\mathbf{b}\|^2 = 1 $ and $ \mathbf{c} \cdot \mathbf{c} = \|\mathbf{c}\|^2 = \frac{4}{7} $. Since $ \mathbf{c} \times \mathbf{a} $ is orthogonal to $ \mathbf{c}, $ \[\mathbf{c} \cdot (\mathbf{c} \times \mathbf{a}) = 0.\]Finally, $ (\mathbf{c} \times \mathbf{a}) \cdot (\mathbf{c} \times \mathbf{a}) = \|\mathbf{c} \times \mathbf{a}\|^2 $. Let $ \theta $ be the angle between $ \mathbf{a} $ and $ \mathbf{c} $. Then \[\|\mathbf{c} \times \mathbf{a}\| = \|\mathbf{a}\| \|\mathbf{c}\| \sin \theta = \frac{2}{\sqrt{7}} \sin \theta,\]so $ \|\mathbf{c} \times \mathbf{a}\|^2 = \frac{4}{7} \sin^2 \theta $. Hence, \[\frac{4}{7} + \frac{4}{7} \sin^2 \theta = 1.\]This leads to \[\sin^2 \theta = \frac{3}{4}.\]so \[\sin \theta = \pm \frac{\sqrt{3}}{2}.\]The smallest possible angle $ \theta $ is then $ \boxed{60^\circ} $. The vectors $ \mathbf{a} = \begin{pmatrix} 1/2 \\ \sqrt{3}/2 \\ 0 \end{pmatrix}, $ $ \mathbf{b} = \begin{pmatrix} 2/\sqrt{7} \\ 0 \\ \sqrt{3/7} \end{pmatrix}, $ and $ \mathbf{c} = \begin{pmatrix} 2/\sqrt{7} \\ 0 \\ 0 \end{pmatrix} $ show that an angle of $ 60^\circ $ is achievable.