1. Find the eccentricity of the ellipse

The ellipse is $\frac{x^2}{18}+\frac{y^2}{4}=1$ Here, $a^2=18,\quad b^2=4$ So, $a=\sqrt{18},\quad b=2$ The eccentricity of an ellipse is $e_1=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{18}}=\sqrt{\frac{14}{18}}=\sqrt{\frac79}=\frac{\sqrt7}{3}$

2. Find the eccentricity of the hyperbola

The hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$ Here, $a^2=9,\quad b^2=4$ For a hyperbola, $e_2=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac49}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}$

3. Use the point $(e_1,e_2)$ on the ellipse

Given that $(e_1,e_2)$ lies on the ellipse $15x^2+3y^2=k$ Substitute $x=e_1=\frac{\sqrt7}{3},\qquad y=e_2=\frac{\sqrt{13}}{3}$ Then $k=15\left(\frac{\sqrt7}{3}\right)^2+3\left(\frac{\sqrt{13}}{3}\right)^2$ $=15\cdot \frac79+3\cdot \frac{13}{9}$ $=\frac{105}{9}+\frac{39}{9}$ $=\frac{144}{9}=16$

4. Compare with the options

Thus, $k=16$ So the correct option is B.