But My Calculator Can't Handle It!

Lemma: let $d$ be a digit and $n$ a positive integer. Then the number consisting of the digit $d$ repeated $n$ times is equal to $\underbrace{dd\cdots d}_{n} = \frac d 9 \cdot 10^n - \frac d 9.$

We apply this to the given number: $A_n = \underbrace{66\cdots6}_{n-1}8 = \underbrace{66\cdots6}_{n} + 2 = \frac 23 \cdot 10^n + \frac 43.$ Square this: $A_n^2 = \left(\frac 23\cdot 10^n + \frac 43\right)^2 = \frac 49\cdot 10^{2n} + \frac{16}9\cdot 10^n + \frac{16}9 \\ = \frac 49\cdot 10^{2n} + 2\cdot 10^n - \frac29\cdot 10^n + 2 - \frac29.$ We use the Lemma again, now in the opposite direction: $A_n^2 = \left(\underbrace{44\cdots4}_{2n} + \frac49\right) + 2\cdot 10^n - \left(\underbrace{22\cdots2}_{n} + \frac29\right) + 2 - \frac29 \\ = \left(\underbrace{44\cdots4}_{n}\underbrace{00\cdots0}_{n} + \underbrace{44\cdots4}_{n}\right) - \underbrace{22\cdots2}_{n} + 2\cdot 10^n + 2 \\ = \underbrace{44\cdots4}_{n}\underbrace{00\cdots0}_{n} + \left(\underbrace{44\cdots4}_{n} - \underbrace{22\cdots2}_{n}\right) + 2\cdot 10^n + 2 \\ = \underbrace{44\cdots4}_{n}\underbrace{00\cdots0}_{n} + \underbrace{22\cdots2}_{n} + 2\cdot 10^n + 2 \\ = \underbrace{44\cdots4}_{n}\underbrace{22\cdots2}_{n} + 2\cdot 10^n + 2 \\ = \underbrace{44\cdots4}_{n-1}6\underbrace{22\cdots2}_{n-1}4.$ The sum of digits is therefore $4(n-1) + 6 + 2(n-1) + 4 = 6n + 4,$ which for $n = 10^5$ results in $\boxed{600\:004}$ .

Here we can consider the pattern followed by the squares of 68.

$68^2=4624$

$668^2=446224$

$6668^2=44462224$

We can notice that the number of 6 in the number is same as number of 2 & 4 in the square of the number. Also there is only one six and one four in the middle and end respectively.

In the question there are $(10^5 -1) 6's$

Therefore we need the sum of:

$4(10^5-1) + 6 + 2(10^5-1) + 4$ =

$\boxed{600004}$

Interestingly, Inline Ruby CAN actually handle it..

a=('6'*99_999+'8').to_i

b = a*a

b.to_s.split('').map{|l|l.to_i}.inject(0){|s,l|s+l}

600004