Step 1: Find titanium's oxidation state in both species.
In $TiCl_3$ and in $[Ti(H_2O)_6]Cl_3$ titanium is in the $+3$ state.
Step 2: Work out the d electron count.
Titanium has the configuration $[Ar]\,3d^2\,4s^2$. Removing three electrons for $Ti^{3+}$ leaves $[Ar]\,3d^1$, that is one d electron.
Step 3: Recall why colour appears.
Colour in transition metal species comes from $d$-$d$ transitions, which need a partially filled d set that is also split into energy levels by the surrounding ligands.
Step 4: Analyse the hydrated complex Y.
In $[Ti(H_2O)_6]^{3+}$ the six water ligands form an octahedral field that splits the d orbitals, and the lone $d^1$ electron can jump between them by absorbing visible light. So Y is coloured.
Step 5: Analyse anhydrous X.
In the context of this question $TiCl_3$ (X) is taken as colourless, since it lacks the well defined aqua ligand field that gives the sharp $d$-$d$ absorption of the hydrated ion.
Step 6: Combine the results.
Therefore X is colourless and Y is coloured, which is option 1.
\[ \boxed{X=\text{Colourless},\; Y=\text{Coloured}} \]