The cross section of the $20 \; \text{m}$ tape has area:

$4^2 \pi- 3^2 \pi = 7\pi \text{cm}^2$

Therefore, the $80 \; \text{m}$ tape should have a cross-section area:

$4 \times 7\pi = 28\pi \; \text{cm}^2$

If $r$ is the outer radius of the $80 \;\text{m}$ roll,

$r^2 \pi- 3^2 \pi = 28\pi \text{cm}^2$ 

$\pi(r^2- 3^2) = 28\pi \text{cm}^2$

$r^2- 3^2 = 28 \text{cm}^2$

$r^2 = 37 \text{cm}^2$

Hence, the outer radius of the $80 \;\text{m}$ roll will be approximately $\sqrt{37} \;\text{cm}$