I kinda did it in a long way, not sure this is the right approach to the problem.
First, I'm going to explain the approach for a small series, let's do that for only 5 terms of the given series.
S1 = 2, S2 = 22, S3 = 222, S4 = 2222, S5 = 22222, S6 = 222222 .... S30 = 222222.....2222
2
22
222
2222
22222
Look closely, for the unit digit [right most] there will be 5 of 2's, So 5 * 2 = 10 and 0 is the unit digit, and 1 is carry forward.
then, for the tenth digit, it will be 2 * 4 = 8 + 1 (from the carry forward of unit digit's calculation)
So, the whole point is to sum the value of multiples of 2's and a single carry forward.
Now, let's consider the whole scenario here,
at most right position you will have 30 of 2's So, 30 * 2 = 60, the position value is 0, and carry forward is 6
then come to one position right, now you have 29 of 2's and a previous carry forward of 6. So, 29 * 2 = 58 + 6 = 64, position value will be 4 and carry forward is now 6.
This may seem tedious calculation, but when you do that on paper it's just simple multiplication and addition. By following the same process 19 times you will reach the 11th position, that will be our answer.