Substituting x=0 into y= 12x+b to get y=b, then the coordinate of point d is (0, b).

The coordinate of point ∫C is (0,8),

Although the quadrilateral ABCO is rectangular,

The ordinate of point e is 8,

Substituting y=8 into y= 12x+b gives 12x+b=8 and x= 16-2b=2(8-b).

The coordinate of point e is (2(8-2b), 8),

∴OD=b,CD=8-b,CE=2(8-b),EH=8，

∵ Rectangular paper ABCO is folded along the straight line Y = 12x+B, so that point C falls on point F on the edge of OA, and the folding seam is DE.

∴df=dc=8-b,fh=ce=2(8-b),∠dfe=∠dce=90，

∴∠DFO+∠EFH=90，

And < dfo+< ODF = 90 degrees,

∴∠ODF=∠EFH，

∴Rt△ODF∽Rt△HFE，

∴OFEH=DFEH=ODFH, which means 8=8? b2(8？ b)=bFH，

∴OF=4,FH=2b，

∫OF+FH = OH = CE，

∴4+2b=2(8-b)，

∴b=3.

So the answer is 3.