$\sqrt{37} \; \text{cm}$. The cross section of the $20 \; \text{m}$ tape has area $\pi(4^2 - 3^2) \text{cm}^2 = 7\pi \text{cm}^2$ Therefore, the $80 \; \text{m}$ tape should have a cross-section area $28\pi \; \text{cm}^2$. Hence, the outer radius of the $80 \;\text{m}$ roll will be approximately $\sqrt{37} \;\text{cm}$

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