Let the average of the number of each row and column be integer denoted by $$x$$
Then the sum of numbers in Row 1 = $$4x$$.
Similarly for row 2 and row 3 the sum will be $$4x$$ each. Hence the sum of all the numbers in the table is $$12x$$.
Let $$N$$ be the number only number that is left out of the table.
Then, (1+2+3+4+5+7+9+10+11+12+13+14+15 -N) = $$12x$$
$$106-N\ =\ 12x$$
Since $$x$$ is integer then $$Remainder\left(\ \frac{\ 106-N}{12}\right)=0$$
$$Remainder\left(\ \frac{\ 10-N}{12}\right)=0$$
Hence N has to be 10 which is left out. And average of each row and column will be $$\ \frac{\ 106-10}{12}=\ \frac{\ 96}{12}=8$$
Upon solving the table will look as follows