1. Given quadratic equation

   The roots $\alpha,\beta$ satisfy $3x^2+\lambda x-1=0.$

   Using Vieta's formulas: $\alpha+\beta=-\frac{\lambda}{3},\qquad \alpha\beta=-\frac{1}{3}.$

2. Use the given condition

   We are told: $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=15.$

   Now, $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}.$

   Since $\alpha\beta=-\frac{1}{3}\implies \alpha^2\beta^2=\frac{1}{9},$ we get

   \implies 9(\alpha^2+\beta^2)=15 \implies \alpha^2+\beta^2=\frac{5}{3}.$$

3. Find $(\alpha+\beta)^2$

   We know $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.$

   Substitute the values:

   =(\alpha+\beta)^2+\frac{2}{3}.$$ Therefore, $$(\alpha+\beta)^2=\frac{5}{3}-\frac{2}{3}=1.$$ Hence, $$\alpha+\beta=\pm 1.$$

4. Find $\alpha^3+\beta^3$

   Use the identity: $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta).$

   Let $s=\alpha+\beta$. Then $s=\pm1$ and $\alpha\beta=-\frac{1}{3}$. So $\alpha^3+\beta^3=s^3-3\left(-\frac{1}{3}\right)s=s^3+s.$

   Since $s=\pm1$, $s^3+s=s+s=2s.$

   Thus, $\alpha^3+\beta^3=\pm 2.$

   Therefore, $(\alpha^3+\beta^3)^2=4.$

5. Compute the required value

   We need $6(\alpha^3+\beta^3)^2=6\cdot 4=24.$

6. Option check

   $24$ corresponds to Option B.

Final Answer: $\boxed{24}$