We have:

\(\overline{ab}=10a+b=b^2 \Rightarrow 10a=b(b-1)\\ \Rightarrow b(b-1) \mathrel{\vdots} 10 \Rightarrow \left [ \begin{array}{l} b \mathrel{\vdots}5\\(b-1)\mathrel{\vdots}5 \end{array} \right.\\ \Rightarrow b\in \{ 5,6\} (we \hskip 0.1cm cannot\hskip 0.1cm have\hskip 0.1cm b=0 \hskip 0.1cm or \hskip 0.1cm b=1 \hskip 0.1cm because \hskip 0.1cm it \hskip 0.1cm leads \hskip 0.1cm to \hskip 0.1cm a=0) \quad (1) \)

We also have:

\(\overline{acbc}=1000a+101c+10b=(\overline{ba})^2=(10b+a)^2=100b^2+20ab+a^2\\ \Rightarrow 1000a +10b(1-10b-2a) - 100c=(a^2-c) \Rightarrow (a^2-c)\mathrel{\vdots}10 \Rightarrow \left [ \begin{array}{l} a^2=c\\ a^2 \mod 10=c \end{array} \right.\\ \Rightarrow (a,c)\in \{(1,1), (2,4), (3,9), (4,6), (5,5),(6,6),(7,9),(8,4),(9,1)\} \quad (2) \)

From\((1)\), we have \(a=2\) if \(b=5\), or \(a=3\) if \(b=6\). Combine with \((2)\), we have: \(\overline{abc} \in \{ 254,369\}\).