For random variables X and Y, is there any onebit variable $Z=f(Y)$, such that $I(X;Z)\geq I(X;Y)/B$ where $B$ is the number of bits to represent $Y$?

$\begingroup$ If X and Y are independent, finding Z is trivial. I think you assume that X and Y are not independent, OK? Also, if X is identical to Y, or X=g(Y), the answer is clear! $\endgroup$ – Shahrooz Janbaz Jan 1 at 12:05

$\begingroup$ We exclude such trivial cases. $\endgroup$ – Saber Saleh Jan 2 at 7:30

$\begingroup$ Your definition of B is ambiguous. For instance, if Y can have 3 different values, is B=2 or B=ln(3) or B=H(Y) ? $\endgroup$ – Arnaud Mégret Feb 22 at 14:26
No. The problem of choosing $Z$ with a rate constraint to maximize $I(X;Z)$ is called the information bottleneck problem and is characterized by the solution to an integral equation.
There is no limit to the complexity of how the information about $X$ appears in $Y$.
Consider that after observing $Y=y$ then $X$ is characterized by the conditional distribution $P_{XY=y}$. To preserve information about $X$ then $Z$ must describe which of these conditional distributions appears. But there is no limit to the amount of complexity in this weighted collection of conditional distributions $(P_{Y=y}\cdot P_{XY=y})_y$.
For a specific example, consider $(X,Y)$ with the jointdistribution:
\begin{equation} P_{X,Y} = \frac{1}{6} \begin{bmatrix} 0,1,1\\ 1,0,1\\ 1,1,0 \end{bmatrix}. \end{equation}
Then one can check any nontrivial $Z$ as you described gives the same $H(Z),\ H(X,Z)$. One can also check $I(X;Z)=\log_2 3\frac{4}{3} < I(X;Y)/H(Y)=1\frac{1}{\log_2 3}$.