Digit $4896$ of the number$122333444455555666666\ldots$ I need find the $4896$th of the number $1223334444555556666667777777\ldots$? 
I have tried to solve the problem using number bases but I failed
 A: I'm assuming after nine $9$'s it continues with ten $10$'s, etc. Note $\frac{9(9+1)}{2}=45$.   
Also, to make it an actual number, it stops at the $45+2\cdot \left(\frac{99(99+1)}{2}-45\right)$'th digit.  
Here $10\le n\le 98$ is such that $n+1$ is the last number that is being repeated when $4896$'th digit is reached.
$$45+2\cdot \left(\frac{n(n+1)}{2}-45\right)\le 4896<45+2\cdot \left(\frac{(n+1)(n+2)}{2}-45\right),$$   
which implies $n=69$ (simple quadratic inequalities, or see WolframAlpha).  
So our answer is either $7$ or $0$. Since $4896$ is even, $\frac{9(9+1)}{2}$ is odd, our digit is $7$.
A: This answer matches the other answers given and should confirm their work, but will perhaps provide an appropriate change of language to make the problem clearer and avoid technicalities caused by trying to define this as a number.  It also, I feel, explains some of the steps in a bit more detail.
Let us define $S(n)$ to be the string generated by concatenating $n$ copies of the number $n$ (in base 10) to the end of the string $S(n-1)$ where $S(0)$ is the empty string.  (Note, here I use the word string to imply that each character in the number $n$ is counted as a separate character).
Our question then is

What is the $4896^{th}$ character in the string $S(n)$ for $n$ large enough to have a $4896^{th}$ character?

It is clear to see that the string $S(n)$ for $1\leq n \leq 9$ is going to be of length $T(n)=\frac{n(n+1)}{2}$.  Note further that for $n$ a two digit number, the length of $S(n)$ is going to be $2n$ longer than $S(n-1)$ as each occurrence of $n$ will occupy two character spaces.
Thus, $\text{length}(S(n)) = \begin{cases}\sum\limits_{i=0}^n i & 0\leq n\leq 9\\ \sum\limits_{i=0}^9 i + \sum\limits_{k=10}^n 2k & 10\leq n\leq 99\\
\vdots\end{cases}$
This can be simplified tremendously however.  Note that by adding and subtracting $\sum\limits_{i=0}^9 i$ to the second case and relabeling the index variable, you get $\text{length}(S(n)) = 2\sum\limits_{i=0}^9 i + 2\sum\limits_{k=10}^n k - \sum\limits_{j=0}^9 j = 2\sum\limits_{k=0}^n k - \sum\limits_{i=0}^9 i =  2T(n)-T(9)= 2T(n)-45$ for $10\leq n\leq 99$.  (Here, $T(n)$ is the $n^{th}$ triangle number, which is given by $\sum\limits_{i=0}^n i = T(n) = \frac{n(n+1)}{2}$)
You get then that $\text{length}(S(69))=4785$ and that $\text{length}(S(70))=4925$.  These two facts imply that since the $4896^{th}$ character was not present in $S(69)$ but instead was newly added to the string during the $70^{th}$ step and so is either a $7$ or a $0$.
The $4896^{th}$ character will be the $111^{th}$ character to be added during the $70^{th}$ cycle (as the $4785^{th}$ character was the final $9$ to be added in the $69^{th}$ cycle and $4896-4785=111$), and thus will correspond to a $7$ (as each odd character being added in the $70^{th}$ cycle will be a $7$ and each even character being added will be a $0$).
Thus, the final answer is $7$.
