The area [tex]a[/tex] of the smaller sector is such that
[tex]\dfrac a{64\pi}=\dfrac{40^\circ}{360^\circ}\implies a=\dfrac{64\pi}9[/tex]
Then the area of the larger sector is
[tex]64\pi-\dfrac{64\pi}9=\dfrac{512\pi}9\approx\boxed{56.9}\pi[/tex]
