$\text{ If in the expansion of }{\left(a^{\frac{1}{3}}+b^{\frac{1}{9}}\right)}^{6561}\text{ ; where }a,b\text{ are distinct prime numbers, the number of irrational terms is }\mathrm{N},$

$\mathrm{then} \mathrm{the} \mathrm{value} \mathrm{of} \frac{\mathrm{N}}{100}\text{ is }$

$\begin{array}{l}\text{ Given expansion is }{\left(a^{\frac{1}{3}}+b^{\frac{1}{9}}\right)}^{6561}\\ T_{\mathrm{r}+1}{=}^{6561}{\mathrm{C}}_{\mathrm{r}}{\left(a^{\frac{1}{3}}\right)}^{6561-\mathrm{r}}\left(b{)}^{\frac{\mathrm{r}}{9}}{=}^{6561}{\mathrm{C}}_{\mathrm{r}}\right(a{)}^{\frac{(6561-\mathrm{r})}{3}}(b{)}^{\frac{\mathrm{r}}{9}}\\ \left(\because \mathrm{In}(x+a{)}^{\mathrm{n}}\text{ expansion },T_{\mathrm{r}+1}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{x}^{\mathrm{r}}{a}^{\mathrm{n}-\mathrm{r}}\right)\\ \text{ Since, }{\mathrm{T}}_{\mathrm{r}+1}\text{ will be integral if }\frac{\mathrm{r}}{9}\text{ and }\frac{6561-\mathrm{r}}{3}\text{ both are integers }\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{r}=0,9,18,\dots \dots ,6561\\ \therefore \text{ Number of values of }\mathrm{r}=\frac{6561}{9}+1=730\\ \text{ So, number of irrational terms }=6562-730=5832\\ \Rightarrow \frac{\mathrm{N}}{100}=\frac{5832}{100}\\ =58.32\end{array}$

Therefore, the correct answer is 58.32