The number under consideration can be expressed as follows:
$$\begin{aligned} & (10^{2n-1} + 10^{2n-2} + \dots + 10^{n+1}) + 2 \cdot 10^n + 8 \cdot (10^{n-1} + 10^{n-2} + \dots + 10^2) + 96 \\ &= 10^{n+1} \cdot \frac{10^{n-1}-1}{9} + 2 \cdot 10^n + 8 \cdot 10^2 \cdot \frac{10^{n-2}-1}{9} + 96 \\ &= \frac{10^{2n} - 10^{n+1} + 18 \cdot 10^n + 800 \cdot 10^{n-2} - 800 + 9 \cdot 96}{9} \\ &= \frac{10^{2n} + 16 \cdot 10^n + 64}{9} = \left( \frac{10^n + 8}{3} \right)^2. \end{aligned}$$
As required, we have obtained a perfect square, because the number $10^n+8$ is divisible by 3, as the sum of its digits equals 9.

$1296 = 36^2, 112896 = 336^2, 11128896 = 3336^2, \dots$
we easily guess that for each $n \ge 2$,
$$\underbrace{1\dots128}_{n-1} \underbrace{896}_{n-2} = \underbrace{33\dots36^2}_{n-1}.$$
The exact proof can be done by using the usual multiplication scheme:
$$\begin{array}{r} 333\dots3336 \\ \times 333\dots3336 \\ \hline 2000\dots0016 \\ 10000\dots008 \\ 100000\dots08 \\ 1000000\dots8 \\ \vdots \\ 1\dots000008 \\ 10\dots00008 \\ 100\dots0008 \\ \hline 111\dots112888\dots8896 \end{array}$$

Both (identical) factors are $n$-digit, hence an $(n + 1)$-digit number stands in each of the $n$ rows between the two delimiting lines. From this fact, it is easy to determine the values of digits (including the numbers of appearances) in the resulting product.