$\because$四边形$ABCD$是矩形，
$\therefore \angle ABC=\angle A=90^{\circ}$，$AD$∥$BC$，
$\because BE$平分$\angle ABC$，
$\therefore \angle ABE=\angle CBE=45^{\circ}$，
$\therefore \angle AEB=\angle CBE=45^{\circ}$，
$\because \angle BEF=15^{\circ}$，
$\therefore \angle AEF=60^{\circ}=\angle CFE$，
$\because CG\bot EF$，
$\therefore \angle CGF=\angle CGE=90^{\circ}$，
$\therefore \angle FCG=30^{\circ}$，
$\therefore FG=\frac{1}{2}CF$，
$\because CF=2BF$，
$\therefore GF=BF$，
$\therefore \angle FBG=\angle FGB$，
$\therefore \angle CFE=2\angle FBG=2\angle FGB$，
$\therefore \angle FBG=\angle FGB=30^{\circ}=\angle FCG$，
$\therefore BG=CG$，
$\because \angle FGB=\angle BEF+\angle EBG$，
$\therefore \angle EBG=15^{\circ}=\angle BEF$，
$\therefore BG=EG=CG$，
$\therefore \triangle EGC$是等腰直角三角形，
$\because CE=6$，
$\therefore CG=\frac{\sqrt{2}}{2}CE=3\sqrt{2}=BG$；
故答案为：$3\sqrt{2}$.

$\because$四边形$ABCD$是矩形，
$\therefore \angle ABC=\angle A=90^{\circ}$，$AD$∥$BC$，
$\because BE$平分$\angle ABC$，
$\therefore \angle ABE=\angle CBE=45^{\circ}$，
$\therefore \angle AEB=\angle CBE=45^{\circ}$，
$\because \angle BEF=15^{\circ}$，
$\therefore \angle AEF=60^{\circ}=\angle CFE$，
$\because CG\bot EF$，
$\therefore \angle CGF=\angle CGE=90^{\circ}$，
$\therefore \angle FCG=30^{\circ}$，
$\therefore FG=\frac{1}{2}CF$，
$\because CF=2BF$，
$\therefore GF=BF$，
$\therefore \angle FBG=\angle FGB$，
$\therefore \angle CFE=2\angle FBG=2\angle FGB$，
$\therefore \angle FBG=\angle FGB=30^{\circ}=\angle FCG$，
$\therefore BG=CG$，
$\because \angle FGB=\angle BEF+\angle EBG$，
$\therefore \angle EBG=15^{\circ}=\angle BEF$，
$\therefore BG=EG=CG$，
$\therefore \triangle EGC$是等腰直角三角形，
$\because CE=6$，
$\therefore CG=\frac{\sqrt{2}}{2}CE=3\sqrt{2}=BG$；
故答案为：$3\sqrt{2}$.