Given, 

${\left(3+6x\right)}^n=3^n{\left(1+2x\right)}^n$

Now if $T_9$ is numerically greatest term, then $T_8\le T_9\ge T_{10}$

So, ${}^{n}C_7{3}^{n-7}{\left(6x\right)}^7\le {}^{n}C_8{3}^{n-8}{\left(6x\right)}^8\ge {}^{n}C_9{3}^{n-9}{\left(6x\right)}^9$

$\Rightarrow \frac{n!}{\left(n-7\right)!7!}9\le \frac{n!}{\left(n-8\right)!8!}3·\left(6x\right)\ge \frac{n!}{\left(n-9\right)!9!}{\left(6x\right)}^2$

$\Rightarrow \underbrace{\frac{9}{\left(n-7\right)\left(n-8\right)}}\le \underbrace{\frac{18\left(\frac{3}{2}\right)}{\left(n-8\right)8}\ge \frac{36}{9.8}\frac{9}{4}}$

$\Rightarrow 72\le 27\left(n-7\right)$ and $27\ge 9\left(n-8\right)$

$\Rightarrow \frac{29}{3}\le n$ and $n\le 11$

So, $n_0=10$

For ${\left(3+6x\right)}^{10}$

Now $T_{r+1}={}^{10}C_r{3}^{10-r}{\left(6x\right)}^r$

For coefficient of $x^6$

$r=6\Rightarrow {}^{10}C_6{3}^4·6^6$

For coefficient of $x^3$

$r=3\Rightarrow {}^{10}C_3{3}^7·6^3$

So, $k=\frac{{}^{10}C_6}{{}^{10}C_3}·\frac{3^4·6^6}{3^7·6^3}=\frac{10!7!3!}{6!4!10!}·8$

$\Rightarrow k=14$

$\therefore k+n_0=24$