Fix a table. Choose a boy and a girl for the table. Choose an order for the boy and girl.
You have six boys and six girls to choose from. Then, you can order them in two ways. So, this is $6\times 6\times 2$ ways to arrange one boy and one girl at the first table.
For the second table, you have five boys left, five girls left, and two ways to arrange then, entirely independent from the arrangement of the first table (the only dependency is that you have one fewer boy and one fewer girl to choose from).
So, there are $5\times 5\times 2$ ways to arrange one boy and one girl at the second table (after a boy and girl were already arranged at the first table).
For the third table, you have $4\times 4\times 2$ ways to arrange them. For the fourth table, you have $3\times 3 \times 2$ ways to arrange them. For the fifth table, you have $2\times 2\times 2$ ways to arrange them. For the last table, you have $1\times 1\times 2$ ways to arrange them.
Because each table is arranged independently from the arrangements of the previous tables, you can apply the product principle. So, the total is:
$$(6\cdot 6\cdot 2)(5\cdot 5\cdot 2)(4\cdot 4\cdot 2)(3\cdot 3\cdot 2)(2\cdot 2\cdot 2)(1\cdot 1\cdot 2) = 2^6(6!)^2$$