At first $X$ is chosen uniformly from $1-100$, $P($choosing $X)=\frac {1}{100}$
Now, $Y$ is chosen from the set$\{1,2....x\}$, so this set should definitely contain atleast $30$ elements from $1-30$, then only $Y=30$ can be chosen, $P(Y=30)=\frac{1}{30}$
We can expand the set to contain $31$ elements, then $P(Y=30)=\frac {1}{31}$
......
We can expand the set till it contains all the $100$ elements, then $P(Y=30)=\frac {1}{100}$
So, $P(Y=30)=\frac{1}{100}*({\frac{1}{30}}+{\frac{1}{31}}+{\frac{1}{32}}+......+{\frac{1}{100}})$
Correct Answer: $B$