If solution set of the equation $3\left[{\log }_{2}x-2\right]-2\left\{{\log }_{2}x\right\}={\log }_{2}x$ is subset of [a, b),where [.] = G.I.F. and {.} represents fractional part,
then [b] - [a] is

Given equation,$3\left[{\log }_{2}x-2\right]-2\left\{{\log }_{2}x\right\}={\log }_{2}x$
can be re-written as 
$\begin{array}{l}3\left[{\log }_{2}x\right]-6-2\left\{{\log }_{2}x\right\}={\log }_{2}x\\ \Rightarrow 3\left({\log }_{2}x-\left\{{\log }_{2}x\right\}\right)-6-2\left\{{\log }_{2}x\right\}={\log }_{2}x\\ \Rightarrow \left\{{\log }_{2}x\right\}=\frac{2{\log }_{2}x-6}{5}\end{array}$
since $0\le \left\{x\right\}<1$, we have
$\begin{array}{l}0\le \frac{2{\log }_{2}x-6}{5}<1\Rightarrow 6\le 2{\log }_{2}x<11\\ \Rightarrow 3\le {\log }_{2}x<\frac{11}{2}\Rightarrow 8\le x<32\sqrt{2}=45.25\dots \end{array}$
Therefore solution of the equation must lie in the interval [8, 45.25...).
$\therefore \left[a\right]=8,\left[b\right]=45,\left[b\right]-\left[a\right]=37$