We have to find the sum of all the digits of the numbers from 10 to 100.

Let's leave aside 100 for now. We are left with 9 sets of 10 numbers each.

10, 11, ..... 19 20, 21, ..... 29 90, 91, ..... 99

Counting Units Digits: Sum of unit digits of each of these 9 sets = $0 +1 +2 + .... + 9 = 9 \times 10 / 2 = 45$

[ Sum of first n natural numbers = $n (n + 1) / 2$ ]

So, sum of all the unit digits of the 9 sets = $45 \times 9 = 405$

Counting Tens Digits: Let's count the tens digits of 10, 20, 30 ..., 90, and then 11, 21, 31, ..., 91, and so on.

Sum of tens digits of 10, 20, 30 ..., 90 = $1 + 2 + 3 + .... + 9 = 9 \times 10 / 2 = 45$

So, sum of all the tens digits = $45 \times 10 = 450$

So, the sum of all the digits of the numbers from 10 to 100 = $405 + 450 + 1 = 856$

Hence, option (B) is correct.