A does 80% of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?

Work done by A in 20 days = $$\frac{80}{100} = \frac{4}{5}$$

Work done by A in 1 day = $$\frac{\frac{4}{5}}{20} = \frac{4}{100} = \frac{1}{25}$$

Work done by A and B in 3 days = $$\frac{20}{100} = \frac{1}{5}$$ (Because remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = $$\frac{1}{15}$$

Work done by B in 1 day = $$\frac{1}{15} - \frac{1}{25} = \frac{2}{75}$$

=> B can complete the work in $$\frac{75}{2}$$ days = 37.5 days