Let for the 9 ^ th term in the binomial expansion of (3 + 6x) ^ n ,...

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, $C7 n {3}^{n - 7}{\left(6x\right)}^{7} \le C8 n {3}^{n - 8}{\left(6x\right)}^{8} \ge C9 n {3}^{n - 9}{\left(6x\right)}^{9}$

$⇒ \dfrac{n !}{\left(n - 7\right) ! 7 !}9 \le \dfrac{n !}{\left(n - 8\right) ! 8 !}3 \cdot \left(6x\right) \ge \dfrac{n !}{\left(n - 9\right) ! 9 !}{\left(6x\right)}^{2}$

$⇒ \mathop{\dfrac{9}{\left(n - 7\right)\left(n - 8\right)}}_{⏟} \le \mathop{\dfrac{18\left(\dfrac{3}{2}\right)}{\left(n - 8\right)8} \ge \dfrac{36}{9 . 8}\dfrac{9}{4}}_{⏟}$

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

$⇒ \dfrac{29}{3} \le n$ and $n \le 11$

So, ${n}_{0} = 10$

For ${\left(3 + 6x\right)}^{10}$

Now ${T}_{r + 1} = Cr 10 {3}^{10 - r}{\left(6x\right)}^{r}$

For coefficient of ${x}^{6}$

$r = 6 ⇒ C6 10 {3}^{4} \cdot {6}^{6}$

For coefficient of ${x}^{3}$

$r = 3 ⇒ C3 10 {3}^{7} \cdot {6}^{3}$

So, $k = \dfrac{C6 10}{C3 10} \cdot \dfrac{{3}^{4} \cdot {6}^{6}}{{3}^{7} \cdot {6}^{3}} = \dfrac{10 ! 7 ! 3 !}{6 ! 4 ! 10 !} \cdot 8$

$⇒ k = 14$

$∴ k + {n}_{0} = 24$