1. General term of the expansion

For $\left(\alpha x^{1/9}+\beta x^{-1/6}\right)^{10},$ the general term is $T_{r+1}=\binom{10}{r}(\alpha x^{1/9})^{10-r}(\beta x^{-1/6})^r.$ So, $T_{r+1}=\binom{10}{r}\alpha^{10-r}\beta^r x^{\frac{10-r}{9}-\frac{r}{6}}.$

2. Find the term independent of $x$

For the constant term, exponent of $x$ must be zero: $\frac{10-r}{9}-\frac{r}{6}=0.$ Multiply by $18$: $2(10-r)-3r=0$ $20-2r-3r=0$ $20-5r=0$ $r=4.$

Hence the constant term is $\binom{10}{4}\alpha^6\beta^4=210\alpha^6\beta^4.$

3. Use the given condition

Given $\alpha^3+\beta^2=4, \qquad \alpha>0,\ \beta>0.$ Let $u=\alpha^3, \qquad v=\beta^2.$ Then $u+v=4,$ and the constant term becomes $210\alpha^6\beta^4=210(\alpha^3)^2(\beta^2)^2=210u^2v^2.$ So we need to maximize $210u^2v^2$ subject to $u+v=4, \quad u,v>0.$

4. Maximize $u^2v^2$

Since $u^2v^2=(uv)^2,$ it is enough to maximize $uv$ subject to $u+v=4$.

By AM-GM, $uv\leq \left(\frac{u+v}{2}\right)^2=\left(\frac{4}{2}\right)^2=4,$ with equality when $u=v=2.$ Therefore, $u^2v^2\leq 4^2=16.$ Hence the maximum constant term is $210\cdot 16=3360.$

Given this maximum is $10k$, $10k=3360 \implies k=336.$

5. Check options

Thus the correct option is: $\boxed{336}$ which is Option B.