To find the probability of $101^{th}$ ball be black can be given as the ratio of number of favourable cases where $101^{th}$ position is black ball to the total cases where $101^{th}$ can be anything (black or white).
$\underline{\textbf{Number of Favourable Cases}}$
We set the $101^{th}$ position ball as black, now we are left with $899$ black balls and $100$ white balls. We can choose $0$ of $899$ black balls and place it on the left of $101^{th}$ position, so to choose $0$ position(s) of $100$ we can do $^{100}C_0$, and for the leftover positions on left we can fill white balls, same thing we can do with leftover black balls (here, $899-0=899$) on right of $101^{th}$ position. Choose $899$ of $899$ positions and set black balls there, and for the rest of the positions we can fill white balls.
One thing to note is, we are not concerned with arrangement of balls as they are identical within their class, hence just selecting the positions to place them will get the job done.
So, when we choosed $0$ black balls, number of favourable cases are $^{100}C_0 \ ^{899}C_{899}$. Just like that if we choose $1$ black ball then we will have $^{100}C_1 \ ^{899}C_{898}$, for $2$ black balls $^{100}C_2 \ ^{899}C_{897}$ and so on.
$$\begin{aligned}\text{Total Favourable cases} &= ^{100}C_0 \ ^{899}C_{899} + ^{100}C_1 \ ^{899}C_{898} + \dots + ^{100}C_{100} \ ^{899}C_{799}\\
&= \sum_{x=0}^{100} \ ^{100}C_x \ ^{899}C_{899-x} \\
&= \sum_{x=0}^{100} \ ^{100}C_{100-x} \ ^{899}C_{x} \\
&= ^{100+899}C_{100} \\
&= ^{999}C_{100}\\\end{aligned}$$
$\sum_{x=0}^{100} \ ^{100}C_x \ ^{899}C_{899-x} = ^{999}C_{100}$ using Vandermonde's identity
$\underline{\textbf{Total Cases}}$
Total cases we can find by selecting $900$ positions to place blacks balls of $1000$ positions. Hence, $^{1000}C_{900}$.
$$Probability = \dfrac{^{999}C_{100}}{^{1000}C_{100}} = \dfrac{9}{10}$$
$\textbf{Option (A) is correct}$