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