THINK ABOUT IT

In order to proceed we need to look at the digit sum of the number and create restraints at each stage of the "trial-and-improvement" method. Here's how it goes....
Let the number be n(0)n(1)n(2)..n(9) such that: $\ n(0)+n(1)+n(2)+n(3)+...+n(7)+n(8)+n(9)$ $\ = n(1)+2n(2)+3n(3)+...+7n(7)+8n(8)+9n(9) \ (1)$ The changing coefficients can be explained with an example. If n(5) = 3 then there are digits equal to 5 giving a "sub-sum" of 5(3) = 15. Note that $\ n(k) \ne k$ otherwise we get a contradiction and $\ n(k) = 0 \implies\ n(0) \ne k$ . Equation (1) gives: $\ n(2)+2n(3)+3n(4)+...+7n(8)+8n(9) = n(0) \leq\ 9$ . $\ n(6)+n(7)+n(8)+n(9) \leq\ 1| \ n(5)+n(6)+n(7)+n(8)+n(9) \leq\ 2$ $\ ... n(3) +n(4)+n(5)+n(6)+n(7)+n(8)+n(9) \leq\ 4$ $\ Case \ n(9) = 1$ : If n(0) = 9, then n(9) = 0 or 1 -> No solution. If n(0) = 8, then n(2) = 1 -> No solution. Therefore, n(9) = 0. $\ Case \ n(9) = 0 \ ,\ n(8) = 1$ : For n(0) = 8 (n(0) = 9 is an immediate contradiction) , so n(2) = 1 -> No solution. If n(0) = 7.

Continuing this method we find that: $\ n(9) = n(8) = n(7) = 0 \ , \ n(6) = 1$ : Equation (1) implies that n(2) = 1, leaving 6n(2)10001000, so b= 2. $\large{\therefore\ answer = 6210001000}$

Going further shows that this solution is unique.