Question 1065051
NOTE: This is a strange problem wording,
and I suspected a typo in the wording you posted.
If so, check the wording carefully and post again.
 
For a square ,
{{{area=side^2}}} <---> {{{side=sqrt(area)}}} .
In this case, for the square in the problem,
{{{side=sqrt(9in^2)=3in}}} .
With the square inscribed in a circle {{{drawing(300,300,-1.5,1.5,-1.5,1.5,
rectangle(-1,-1,1,1),
circle(0,0,sqrt(2)),red(triangle(-1,-1,1,-1,1,1)),
red(rectangle(0.9,-1,1,-0.9)),
locate(-0.2,-1,red(3in)),locate(1.05,0.1,red(3in)),
locate(0,0,red(d))
)}}} ,
the diameter, {{{red(d)}}} , of the circle
is the diagonal, {{{red(d)}}} , of the square.
According to the Pythagorean theorem,
{{{red(d)^2=red(3in)^2+red(3in)^2=2*9in^2}}} ,
so
{{{d=sqrt(2*9in^2)=sqrt(2)*sqrt(9)}}}{{{in=3sqrt(2)}}}{{{in}}} .

For any circle, {{{radius=diameter/2}}} ,
{{{area=pi*radius^2}}} , and
circumference=pi*diameter=pi*radius/2}}} .
For any circle,
the ratio of circumference of the circle to the area of the circle is
{{{(pi*radius/2)/(pi*radius^2)=2/radius}}} .
For the circle in the problem,
{{{radius=3sqrt(2)/2}}}{{{in}}} ,
and we could calculate that ratio as
{{{2/radius}}}{{{"="}}}{{{2/((3sqrt(2)/2))}}}{{{"="}}}{{{4/3sqrt(2)}}}{{{"="}}}{{{2sqrt(2)/3}}} ,
with units of {{{1/in}}} or {{{in^(-1)}}} .
Maybe you were expected to calculate
{{{circumference=3sqrt(2)*pi}}}{{{in}}} ,
{{{radius=3sqrt(2)/2}}}{{{in}}} ,
{{{area=pi*(3sqrt(2)/2)^2}}}{{{in^2=pi*9*2/4}}}{{{in^2=9pi/2}}}{{{in^2}}} ,
and then further calculate the ratio as
{{{3sqrt(2)*pi/((9pi/2))}}}{{{"="}}}{{{2sqrt(2)/3}}} ,
with units of {{{1/in}}} or {{{in^(-1)}}} .