You divide the number of physical qubits by 5 to get the number of fault tolerant error corrected qubits. [0] If their algorithm works, they need a 1860 (372*5) qubit computer to break 2048 bit RSA. IBM expects to get there by 2025. [1] [0] https://en.wikipedia.org/wiki/Five-qubit_error_correcting_co... [1] https://www.ibm.com/quantum/roadmap
For example, using the surface code, a back of the envelope estimate would be that you need a code distance of d = ln(number_of_operations). Each logical qubit will use 2d^2 physical qubits. So for a million operations you'd need around 400 physical qubits per logical qubit and for a trillion operations you'd need around 1500 physical qubits per logical qubit. So, way more than 5.
(A major practical obstacle to using almost-anything-that-isn't-the-surface-code is that the surface code has forgiving connectivity and maximum-allowed-physical-noise requirements.)