Let the equation of two ellipses be ${\mathrm{E}}_1:\frac{{\mathrm{x}}^2}{3}+\frac{{\mathrm{y}}^2}{2}=1$ and ${\mathrm{E}}_2:\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1.$ If the product of their eccentricities is  1/2, then the length of the minor axis of ${\mathrm{E}}_2$ is

Given  ${\mathrm{E}}_1 : \frac{{\mathrm{x}}^2}{3}+\frac{{\mathrm{y}}^2}{2}=1 (\mathrm{a}>\mathrm{b})$

   $$\begin{gathered} {\mathrm{e}}_1=\sqrt{\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{\frac{3-2}{3}}=\frac{1}{\sqrt{3}} \\ {\mathrm{E}}_2 : \frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ {\mathrm{e}}_2=\sqrt{\frac{16-{\mathrm{b}}^2}{16}}=\frac{\sqrt{16-{\mathrm{b}}^2}}{4} \end{gathered}$$

Given that ${\mathrm{e}}_1.{\mathrm{e}}_2=\frac{1}{2}$

$$\begin{gathered} \Rightarrow \frac{1}{\sqrt{3}} \left(\frac{\sqrt{16-{\mathrm{b}}^2}}{4}\right)=\frac{1}{2} \\ \Rightarrow \sqrt{16-{\mathrm{b}}^2}=2\sqrt{3} \end{gathered}$$

squaring on both sides

$$\begin{gathered} \Rightarrow 16-{\mathrm{b}}^2=12 \\ \Rightarrow {\mathrm{b}}^2=4 \\ \Rightarrow \mathrm{b}=2 \end{gathered}$$

Now length of minor axis of ${\mathrm{E}}_2=2\mathrm{b}=4$