$$Aluminium Chloride in acidified aqueous solution forms octahedral ${\left[\mathrm{Al}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}$ ion.

${\mathrm{AlCl}}_3{\stackrel{}{\stackrel{\text{Acidified aqueous solution}}{\to }{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}}}^{}$

In this complex the $3\mathrm{d}$ orbital of Al are involved and the hybridisation state of Al is ${\mathrm{sp}}^3{\mathrm{d}}^2$.

Hence, the answer is option A.