Let $a=\sqrt[3]{2}=2^{1/3},\qquad b=\frac{1}{\sqrt[3]{3}}=3^{-1/3}.$ Then the expansion is $(a+b)^n.$

General term from the beginning: The $(r+1)^{\text{th}}$ term is $T_{r+1}=\binom{n}{r}a^{n-r}b^r.$ So the $15^{\text{th}}$ term from the beginning is obtained by taking $r=14$: $T_{15}=\binom{n}{14}a^{n-14}b^{14}.$

The $15^{\text{th}}$ term from the end: In the expansion of $(a+b)^n$, the $(k)^{\text{th}}$ term from the end is the $(n-k+2)^{\text{th}}$ term from the beginning. Hence the $15^{\text{th}}$ term from the end is the $(n-13)^{\text{th}}$ term from the beginning. So here $r=n-14$, and $T_{15}^{(\text{end})}=\binom{n}{n-14}a^{14}b^{n-14}=\binom{n}{14}a^{14}b^{n-14}.$

Given ratio: $\frac{T_{15}}{T_{15}^{(\text{end})}}=\frac{1}{6}.$ Substitute the terms: $\frac{\binom{n}{14}a^{n-14}b^{14}}{\binom{n}{14}a^{14}b^{n-14}}=\frac{1}{6}.$ The binomial coefficients cancel: $\left(\frac{a}{b}\right)^{n-28}=\frac{1}{6}.$

Compute $\dfrac{a}{b}$: $\frac{a}{b}=\frac{2^{1/3}}{3^{-1/3}}=(2\cdot 3)^{1/3}=6^{1/3}.$ Therefore, $\left(6^{1/3}\right)^{n-28}=\frac{1}{6}=6^{-1}.$ So, $6^{(n-28)/3}=6^{-1}.$ Equating powers, $\frac{n-28}{3}=-1.$ Hence, $n-28=-3 \Rightarrow n=25.$

Now calculate ${}^nC_3$: $\binom{25}{3}=\frac{25\cdot24\cdot23}{3\cdot2\cdot1}.$ $\binom{25}{3}=\frac{25\cdot24\cdot23}{6}=25\cdot4\cdot23=2300.$

Compare with options: Option B is $2300.$