If the design width doubles, then I'm assuming that the surface area of the balloon doubles when the balloon is inflated a second time. And the balloon can be considered as a sphere.
Let's let the original radius = r
Then, the original surface area of the balloon = 4pi *r^2
Then, for the suface area to double, the radius must increase by a factor of √2
Proof : 4pi (√2 r)^2 = 4pi (2)r^2 = 8pi*r^2 = twice the original surface area
So...let's find the original radius using the volume for a sphere
71 =(4/3)pi*r^3 divide both sides by (4/3)pi
71/ [ (4/3)pi] = r^3 take the cube root of both sides
cube root ( 71/ [ (4/3)pi] ) = r = about 2.5687cm^3
So......if the original radius increases by a factor of √2, the new volume is :
V = (4/3)pi (2.5687√2)^3 = about 200.8 cm^3
