A and B can do a piece of work in 20 days and B in 15 days.

It is given that A can finish the work in 20 days and B can finish the same work in 15 days.

$\therefore$ Work done by A in 1 day $=\frac{1}{20}$

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

$\therefore$ Work done by $(\mathrm{A}+\mathrm{B})$ in 1 day $=\frac{1}{20}+\frac{1}{15}$

$=\frac{3+4}{60}=\frac{7}{60}$

$\therefore$ Work done by $(\mathrm{A}+\mathrm{B})$ in 2 days $=\frac{14}{60}=\frac{7}{30}$

Remaining work $=1-\frac{7}{30}=\frac{23}{30}$

It is given that the remaining work is done by B.

$\because$ Complete work is done by B in 15 days.

$\therefore \frac{23}{30}$ of the work will be done by $\mathrm{B}$ in $\left(15 \times \frac{23}{30}\right)$ days or $\frac{23}{2}$ days or $11 \frac{1}{2}$ days.

Thus, the remaining work is done by B in $11 \frac{1}{2}$ days.