Correct option: C.$Tl^+$ is more stable than $Al^+$
c
$(i)\,\mathop {A{l^{3 + }}}\limits_{Most\,\,stable} \,\xrightarrow{{{E^o}\, = \, - \,1.66}}Al\,\xleftarrow{{{E^o}\, = \, + \,0.55}}\mathop {A{l^ + }}\limits_{Less\,\,stable} $
$(ii)\mathop {T{l^{3 + }}}\limits_{Less\,\,stable} \,\xrightarrow{{{E^o}\, = \, - + 1.26}}Tl\,\xleftarrow{{{E^o}\, = \, - 0.34}}\mathop {T{l^ + }}\limits_{Most\,\,\,stable} $
$Tl^+$ has negative electrode potential $(E^o = - 0.34)$ means, it does not prefer to convert into $Tl$ but reverse must be preferred that's why it is more stable than $Tl^{3+}$ $(E^o\,=\,+\,1.26)$ In $Al$, $Al^{3+}$ is more stable $E^o=\,-\,1.66$ than $Al^+$ $(E^o\,=\,+0.55)$ and also from $Tl^+$ due to more negative value of $E^o$. Therefore, by comparison it confirms that $Tl^+$ is more stable than $Al^+$.