2 X 2 = 3

We are given that $\overline{TWO}^2 = \overline{THREE}$ .

This tells us that $\overline{TWO} < \sqrt{100000} = 316.2$ , therefore $T$ is $1$ , $2$ or $3$ , It is obvious that $T = \boxed{1}$ because:

• if $T=2$ then the $T$ of $\overline{THREE}$ will be either $4$ or $5 \ne T$
• if $T=3$ then the $T$ of $\overline{THREE}$ will be $9 \ne T$

Therefore, $\overline{THREE} < 20000 \quad \Rightarrow \overline{TWO} < \sqrt{20000} = 141.4$

We also note that for the last digits $O \times O = 10c_1 + E$ , where $c_1 = 0,1,2,3,4,6,8$ is the carried-forward value, $O \ne 0,1,5,6$ else $O=E$ which is unacceptable.

Working out the possible cases as follows:

\[\begin{array} {} 124^2=15376\\ 127^2=16129\\ 128^2=16384\\ 129^2=16641\\ 132^2=17424\\ 134^2=17956\\ 137^2=18769\\ 138^2=19044\\ 139^2=19321 \end{array}\]

We note that there is only one solution: $\overline{TWO} = 138$ and $\overline{THREE} = 19044$

$\Rightarrow \overline{WHO} = \boxed{398}$