Let the three children be $A,B,C$ and the $20$ oranges be distinct.

Since each orange is distinct, for every orange we choose which child receives it.

Total unrestricted distributions: Each of the $20$ oranges can be given to any one of the $3$ children. Hence, $3^{20}$ total distributions.

Subtract distributions where at least one child gets no orange using inclusion-exclusion.

• If child $A$ gets no orange, then each orange goes to either $B$ or $C$: $2^{20}$ ways.
• Similarly for $B$ or $C$.

So total for one child excluded: $\binom{3}{1}2^{20} = 3\cdot 2^{20}.$

Add back distributions where two children get no orange:

• If $A$ and $B$ get no orange, then all oranges go to $C$: $1^{20}=1$ way.
• Similarly for the other pairs.

So total for two children excluded: $\binom{3}{2}1^{20} = 3.$

Therefore, the required number is $3^{20} - \binom{3}{1}2^{20} + \binom{3}{2}1^{20}.$

Now compute: $3^{20} = 3486784401,$ $2^{20} = 1048576.$

Hence, $3\cdot 2^{20} = 3145728.$

So, $3486784401 - 3145728 + 3 = 3483638676.$

Thus the number of ways is $\boxed{3483638676}.$