Here we are having a total of $1000$ Balls, out of which we firstly draw $100$ balls , and then $101^{st}$ ball..
Firstly we have to find expected number of white and black balls in drawn $100$ balls , as both can occur in 100 balls..
We have a situation like this:
Expected number of white balls =$n\times \dfrac{W}{N} = 100\times \left(\dfrac{100}{1000}\right) = 10$
Expected number of black balls =$n\times \dfrac{B}{N} = 100\times \left(\dfrac{900}{1000}\right) = 90$
So, we have drawn $100$ balls($90$ black, $10$ white)
Left balls = ($810$ Black , $90$ White) = $900$ total
Now,
probability for ${101}^{st}$^{ }ball to be black = $\dfrac{810}{900} =\dfrac{9}{10}.$
So, option (A) is Correct