Given, $b\text{ boys and g girls}$

Also given total ways of selecting $3\text{ boys and 2 girls}$ which will be ${}^{b}C_3\times {}^{g}C_2=168$

$\Rightarrow \frac{b\left(b-1\right)\left(b-2\right)}{6}\times \frac{g\left(g-1\right)}{2}=168$

$\Rightarrow b\left(b-1\right)\left(b-2\right)g\left(g-1\right)=7\times 6\times 4\times 3\times 2\times 2$

$\Rightarrow b\left(b-1\right)\left(b-2\right)g\left(g-1\right)=8\times 7\times 6\times 3\times 2$

Now on comparing both side we get, $b=8$ and $g=3 \Rightarrow b+3g=8+9=17$