I cannot attach an image.But follow the steps one by one.
Draw a triangle APQ.angle PQA=90.PQ=h.angle PAQ=45.Say AQ=l.
h/l=tan45
Or,l=h
Now draw a point inside the triangle.This point is B.join AB.AB=d.Drop a perpendicular from B on AQ.name the touching point M.This will from another triangle ABM.angle BAM=30.say AM=x and BM=y.say BN=x'.
SO l=x'+x
Or,h=x'+x
We can find out x from triamgle ABM.
cos30=x/d
Or,x=root(3)d/2
Sin30=y/d
Or,y=d/2
Join PB.drop a perpendicular from B on PQ.name the touching point N.angle PBN=60.NQ=y.PN=h-y.
tan60=(h-y)/x'
x'=(h-d/2)/root3
Put these values in l=h=x+x' and you will get answer.
 
						

							
Typing here is really painful.I typed at one place,it appeared at another place.
After "BM=y" please do the drawing "join PB".then go back to BN=x' and  do each step.
 
 
						

							
In ΔAPQ, Q is right angle, PAQ=π/4, PQ=h (given)
PQ/AQ=tan(π/4)=1 => AQ=PQ=h
Locate point B in this Δ such that BAQ= π/6
From B, drop perpendiculars onto PQ & AQ respectively cutting them at R & C.
Thus ABC & PBR are two Δs with AB=d & BCQR is a rectangle.
 
In triangle ΔABC, AC=d*cos(π/6) = d√3/2, BC = d*sin(π/6)=d/2

Also, BR=CQ=AQ-AC=(h-d√3/2)=(2h-d√3)/2
QR=BC=d/2; PR=PQ-QR=(h-d/2)=(2h-d)/2
PBR = π/3 (given)
 
In ΔPBR, PR/BR=tan(π/3)
=> [(2h-d)/2]/[(2h- d√3)/2]= √3
=> (2h-d)/(2h- d√3)= √3
=> 2h-d= √3*(2h- d√3)
=> 2h-d = (2√3h-3d)
=> 2h*(√3-1)=3d-d=2d
=> d= h*(√3-1)
 
Trust this helps.