Good looking problem!

Lets see, that the first number that work is 1. Now let find a greater one digit number. We can see the numbers are in the form $k(kkk...kkk)$ with the second $k`s$ written $k$ times. See $k(kkk...kkk)={ k }^{ 2 }(111...111)$ with the $1`s$ written $k$ times. So, if $k^{ 2 }(111...111)$ is a square, then, $(111...111)$ is a square.

See $111...111=1+10+...+1{ 0 }^{ k-1 }=\frac { { 10 }^{ k }-1 }{ 9 }$ ; so $\frac { { 10 }^{ k }-1 }{ 9 }$ is a square if $1{ 0 }^{ k }-1$ is also a square.

$1{ 0 }^{ k }-1=999...999$ with the $9`s$ written $k$ times. After seeing the quadratic residues $mod 10$ we know that the units digits is $3$ or $7$ . Now, we check the residues $mod 100$ with the units digits $3$ and $7$ and it should be 99.

${ 7 }^{ 2 }\equiv 49,\quad { 17 }^{ 2 }\equiv 89,\quad { 27 }^{ 2 }\equiv 29,\quad { 37 }^{ 2 }\equiv 69,\quad { 47 }^{ 2 }\equiv 9,\quad { 57 }^{ 2 }\equiv 49,\quad { 67 }^{ 2 }\equiv 89,\quad { 77 }^{ 2 }\equiv 29,\quad { 87 }^{ 2 }\equiv 69,\quad { 97 }^{ 2 }\equiv 9 mod 100$ so $7$ doesn`t works.

${ 3 }^{ 2 }\equiv 9,\quad { 13 }^{ 2 }\equiv 69,\quad { 23 }^{ 2 }\equiv 29,\quad { 33 }^{ 2 }\equiv 89,\quad { 43 }^{ 2 }\equiv 49,\quad { 53 }^{ 2 }\equiv 9,\quad { 63 }^{ 2 }\equiv 69,\quad { 73 }^{ 2 }\equiv 29,\quad { 83 }^{ 2 }\equiv 89,\quad { 93 }^{ 2 }\equiv 49\quad mod\quad 100$ . So, $999...999$ isn`t a square.

Now, let find a greater number , say $N$ is a number with $d$ digits. So, after some reasoning we can see that the first expression is ${ N }^{ 2 }({ 10 }^{ d-1 }+{ 10 }^{ 2(d-1) }+...+{ 10 }^{ x(d-1) })$ . So, as ${ N }^{ 2 }$ is a square, then $({ 10 }^{ d-1 }+{ 10 }^{ 2(d-1) }+...+{ 10 }^{ x(d-1) })$ must be a square. Now, see ${ 10 }^{ d-1 }+{ 10 }^{ 2(d-1) }+...+{ 10 }^{ x(d-1) }={ 10 }^{ d-1 }(1+10+...+{ 10 }^{ x })$ , that is why it was important to see that $1+10+...+{ 10 }^{ x }$ isn`t a square.

Now, we have two cases:

$1)$ If, $d$ is odd, then clearly the expression isn`t a square.

$2)$ If $d$ is even then we can say that ${ 10 }^{ d-1 }(1+10+...+{ 10 }^{ x })=10*{ 10 }^{ 2 m}(1+10+...+{ 10 }^{ x })$ . So the expression is a square if and only if $10(1+10+...+{ 10 }^{ x })$ but that is not possible because $10$ and $1+10+...+{ 10 }^{ x }$ .

So, the only answer is $\boxed { 1 }$ .

Only $1×1$ is a perfect square. So, the answer is $\color{#69047E}{\boxed{1}}$ term is a perfect square.

that's very nice