$(3+6 x )^{ n }={ }^{ n } C _{0} 3^{ n }+{ }^{ n } C _{1} 3^{ n -1}(6 x )^{ I }+\ldots$

$T _{ r +1}{ }^{ n } C _{ r } 3^{ n – r } \cdot(6 x ) r $

$={ }^{ n } C _{ r } 3^{ n – r } \cdot 6^{ I } \cdot x ^{ r } 3^{ n – r } \cdot 3^{ r } \cdot 2^{ r } \cdot\left(\frac{3}{2}\right)^{ r }$

$={ }^{ n } C _{ r } 3^{ n } \cdot 3^{ r } \quad \text { [for } x =\frac{3}{2}]$

$T_{9}$ is greatest of $x =\frac{3}{2}$

So, $T _{9}> T _{10}$ and $T _{9}> T _{8}$

(concept of numerically greatest term)

Here, $\frac{T_{9}}{T_{10}}>1$ and $\frac{T_{9}}{T_{8}}>1$

$\frac{{ }^{n} C_{8} 3^{n} \cdot 3^{8}}{{ }^{n} C_{9} 3^{n} \cdot 3^{9}}>1$ and $\frac{{ }^{n} C_{8} 3^{n} \cdot 3^{8}}{{ }^{n} C_{7} 3^{n} \cdot 3^{7}}>1$

and $\frac{{ }^{ n } C _{8}}{{ }^{ n } C _{7}}>\frac{1}{3}$

and $\frac{ n -7}{8}>\frac{1}{3}$

$\frac{29}{3}< n <11 \Rightarrow n =10= n _{0}$

So, in $(3+6 x )^{ n }$ for $n = n _{0}=10$

i.e., in $(3+6 x )^{10}$, here $T _{ r +1}={ }^{10} C _{ r } 3^{10- r } 6^{ T } x ^{ r }$

$T _{7}={ }^{10} C _{6} 3^{4} \cdot 6^{6} \cdot x ^{6}=210 \cdot 3^{10} \cdot 2^{6} x ^{6}$

$T _{4}={ }^{10} C _{3} 3^{7} 6^{3} x ^{3}=120.3^{10} \cdot 2^{3} x ^{3}$

Ratio of coefficient of $x ^{6}$ and coefficient of $x ^{3}= k$ $\therefore k =\frac{210 \cdot 3^{10} 2^{6}}{120 \cdot 3^{10} \cdot 2^{3}}=\frac{7}{4} \times 2^{3}=14$

So, $k + n _{0}=14+10=24$