Question 412212
<pre><font face = "consolas" color = "indigo" size = 4><b>

{{{drawing(400,400,-3,3,-3,3,

rectangle(-2,-2,2,2), circle(-1,-1,1),circle(-1,1,1),circle(1,1,1),
circle(1,-1,1),circle(0,0,sqrt(2)-1),

circle(-1,-1,.03),circle(-1,1,.03),circle(1,1,.03),
circle(1,-1,.03),circle(0,0,.03) 


 )}}}

Since the area of the square is 16 square units, each side of the 
square is the square root of 16 or 4 units.

This green line: 

{{{drawing(400,400,-3,3,-3,3,

rectangle(-2,-2,2,2), circle(-1,-1,1),circle(-1,1,1),circle(1,1,1),
circle(1,-1,1),circle(0,0,sqrt(2)-1),
green(line(-1,-2,-1,2)),
circle(-1,-1,.03),circle(-1,1,.03),circle(1,1,.03),
circle(1,-1,.03),circle(0,0,.03) 


 )}}}

is also 4 units and goes across both circles on the left.
so each of the four larger circles has radius 1.

Now let's draw a diagonal of the square and letter some points:

{{{drawing(400,400,-3,3,-3,3,
locate(-2,-2,A), locate(-1,-1,C),locate(1-sqrt(2)/2+.06,1-sqrt(2)/2+.11,F),locate(0,0,E),locate(-1+sqrt(2)/2-.05,-1+sqrt(2)/2-.1,D),locate(1,1,G), locate(2.05,2.05,H),

rectangle(-2,-2,2,2), circle(-1,-1,1),circle(-1,1,1),circle(1,1,1),
circle(1,-1,1),circle(0,0,sqrt(2)-1),
green(line(-1,-2,-1,2)), locate(-1,-2,B),
circle(-1,-1,.03),circle(-1,1,.03),circle(1,1,.03),
circle(1,-1,.03),circle(0,0,.03),
red(line(-2,-2,2,2)), 
locate(2,-2,I)

 )}}}


ABC is a right triangle,

AB=1, BC=1, so by the Pythagorean theorem AC = {{{sqrt(2)}}}

and since GH = AC, GH = {{{sqrt(2)}}}

CD and FG are both radii of the larger circles so they are 1 each.

Triangle AIH is a right triangle, AI = 4, IH = 4, so by the
Pythagorean theorem, diagonal AH = {{{4sqrt(2)}}}

We add up the parts of the diagonal AH and equate the sum to {{{4sqrt(2)}}} 

AC + CD + DF + FG + GH = {{{4sqrt(2)}}} 

Substituting the values for the parts that we know the lengths of:

{{{sqrt(2)}}}+1+DF+1+{{{sqrt(2)}}} = {{{4sqrt(2)}}}

Now we solve for DF

{{{2sqrt(2)}}}+2+DF = {{{4sqrt(2)}}}

DF = {{{2sqrt(2)-2}}} 

DF is the diameter of small circle E, so its radius

is {{{expr(1/2)(2sqrt(2)-2)}}} or  

{{{sqrt(2)-1}}}

Edwin</pre>