The distance between the parallel lines $y=2x+4$ and $6x=3y+5,$ is

$2x-y+4=0$ and $6x-\beta y-5=0$

$\Rightarrow 2x-y+4=0$ and $2x-y-\frac{5}{3}=0$

The distance between these two parallel lines is given by

$d=\left|\frac{4+\frac{5}{3}}{\sqrt{2^2+(-1{)}^2}}\right|=\frac{17\sqrt{5}}{15}$