Let two particular boys as one boy, we have only four boys which can be seated at a round table in $3!$ ways.
The two boys together can be arranged in $2$ ways.
So, boys can be seated in $2\times 3!$ ways.

$B_1{B}_2$ are together. Now $4$ girls can be seated at four places (marked $\times$) in $4!$ ways.
$\Rightarrow$ Required number of ways $=3!\times 2\times 4!$
$=6\times 2\times 24=288$ ways