Given,

$7$ boys and $5$ girls are to be seated around a circular such that no two girls to be seated together,

Now we know that $n$ objects can be arranged in a circle in $(n-1)!$ ways.

Let us first arrange $7$ boys in circular arrangement in $\left(7-1\right)!$ ways.

Now there will be $7$ gaps.

So let us select any $5$ gaps out of $7$ gaps and arrange $5$ girls in the chosen gaps. This can be done in ${}^{7}C_5\times 5!$ ways.

Hence, required arrangements are $6!\times {}^{7}C_5\times 5!$

$=6\times 5!\times \frac{7\times 6}{2}\times 5!$

$=126{\left(5!\right)}^2$.

Therefore, required arrangements are $126{\left(5!\right)}^2$