Given that in #\Delta ABC#, #A=\pi/8#, #B={\pi}/3#

#C=\pi-A-B#

#=\pi-\pi/8-{\pi}/3#

#={13\pi}/24#

from sine in #\Delta ABC#, we have

#\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}#

#\frac{a}{\sin(\pi/8)}=\frac{b}{\sin ({\pi}/3)}=\frac{c}{\sin ({13\pi}/24)}=k\ \text{let}#

#a=k\sin(\pi/8)=0.383k#

# b=k\sin({\pi}/3)=0.866k#

#c=k\sin({13\pi}/24)=0.991k#

#s=\frac{a+b+c}{2}#

#=\frac{0.383k+0.866k+0.991k}{2}=1.12k#

Area of #\Delta ABC# from Hero's formula

#\Delta=\sqrt{s(s-a)(s-b)(s-c)}#

#18=\sqrt{1.12k(1.12k-0.383k)(1.12k-0.866k)(1.12k-0.991k)}#

#18=0.1644k^2#

#k^2=109.4506#

Now, the in-radius (#r#) of #\Delta ABC#

#r=\frac{\Delta}{s}#

#r=\frac{18}{1.12k}#

Hence, the area of inscribed circle of #\Delta ABC#

#=\pi r^2#

#=\pi (18/{1.12k})^2#

#=\frac{324\pi}{1.2544k^2}#

#=\frac{811.4445}{109.4506}\quad (\because k^2=109.4506)#

#=7.414\ \text{unit}^2#