2(A + B + C)'s 1 day's work = \(\left(\frac{1}{30}+\frac{1}{24}+\frac{1}{20}\right) =\frac{15}{120}=\frac{1}{8}.\)

Therefore, (A + B + C)'s 1 day's work =  \(\frac{1}{2\times8}=\frac{1}{16}\)

Work done by A, B, C in 10 days =  \(\frac{10}{16}=\frac{5}{8}\)

Remaining work =  \(\left(1-\frac{5}{8}\right)=\frac{3}{8}\)

A's 1 day's work = \(\left(\frac{1}{16}-\frac{1}{24}\right)=\frac{1}{48}\)

Now,   \(\frac{1}{48}\)  work is done by A in 1 day.

So,   \(\frac{3}{8}\)    work will be done by A in     \(\left(48\times\frac{3}{8}\right)\)    =  18 days.