Let the required number of days be $n$ . Then

$\dfrac {n}{20}+\dfrac {n-3}{40}=1-\dfrac {10}{100}\implies n=\boxed {13}$ .

Getting to the answer

Let the whole work be $1$ units (I don't remember the unit of work).

Anil alone can do a job in $20$ days

$\implies$ Work done by Anil in $1$ day = $\dfrac{1}{20}$

Sunil alone can do it in $40$ days

$\implies$ Work done by Sunil in $1$ day = $\dfrac{1}{40}$

Bimal has done $10\%$ of the job

$\implies$ Work done by Bimal in all the time he worked = $10 \% \text{of} 1 =\dfrac{1}{10}= \dfrac{2}{20}$ (for simpler calculation)

Calculations

First, Anil works for $3$ days.

$\implies$ Work done by Anil in $3$ days = $\dfrac{3}{20}$

Work done by Bimal in all the time he worked = $\dfrac{2}{20}$

$\implies$ Work done by Anil in $3$ days and Bimal in all the time he worked = $\dfrac{3}{20} + \dfrac{2}{20} = \dfrac{5}{20} = \dfrac{1}{4}$

$\implies$ Work remaining to be calculated = $\dfrac{3}{4}$

Anil and Sunil both work together for $x$ days.

Work done by Anil and Sunil in $1$ day = $\dfrac{1}{20} + \dfrac{1}{40} = \dfrac{3}{40}$

Work left $=\dfrac{3}{4}$

$x = \dfrac{3}{4} \div \dfrac{3}{40} = 10$

The answer...

Total no. of days $=3$ (for which Alice alone worked) $+ x$

Total no. of days $=3 + 10 =\textcolor{#D61F06}{\boxed{13}}$