The lanthanide ion that would show colour is:

The elements or ions having unpaired electrons show colour. The valence shell electronic configuration of ${\mathrm{La}}^{3+}$ and ${\mathrm{Lu}}^{3+}$ are $4{\mathrm{f}}^0 \text{and} 4{\mathrm{f}}^{14}$ respectively. Hence, they are colourless. ${\mathrm{Gd}}^{3+}$ is also colourless, this is because, here $4\mathrm{f}$ orbitals are half filled,hence stable.As such $\mathrm{f}-\mathrm{f}$ transitions are not possible because they will destroy the symmetry of $4\mathrm{f}$ and make them unstable. ${\mathrm{Sm}}^{3+}$ has the valence shell electronic configuration $4{\mathrm{f}}^5$, hence, it is coloured due to the presence of unpaired electrons.

$\mathrm{L}{\mathrm{a}}^{+3}=$ Colourless
$\mathrm{G}{\mathrm{d}}^{+3}=$ Colourless
$\mathrm{S}{\mathrm{m}}^{+3}=$ Pale yellow colour
$\mathrm{L}{\mathrm{u}}^{+3}=$ Colourless