Let the size of X data be $n$.
So,
$c \log_{2} n = 12$
for some constant $c > 0$.
Now, we know that Y has a data of $32$ elements and gets the output in $20 s$, so,
$c \log_{2} 32 = 20$
which gives: $c = 4$.
So, putting value of $c$ in first equation,
$4 \log_{2} n = 12$
$\Rightarrow \log_{2} n = \frac {12}{4}$
$\Rightarrow n = 2^{3}$
$\Rightarrow n = 8$.
$Answer = 8.$