Let the equations of two ellipses be
$$E_1 : \frac{x^2}{3} + \frac{y^2}{2} = 1$$ and $$E_2 : \frac{x^2}{16} + \frac{y^2}{b^2} = 1$$,
If the product of their eccentricities is $$\frac{1}{2}$$, then the length of the minor axis of ellipse $$E_2$$ is :

First, we write the general form of an ellipse centred at the origin:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$

where $$a$$ is the semi-major axis, $$b$$ is the semi-minor axis and $$a>b>0$$. For such an ellipse the eccentricity $$e$$ is defined by the formula

$$e=\sqrt{1-\frac{b^2}{a^2}}.$$

For the first ellipse $$E_1$$ we have

$$\frac{x^2}{3}+\frac{y^2}{2}=1.$$

Comparing with the standard form, we identify

$$a_1^2=3,\qquad b_1^2=2,$$

and since $$3>2$$, the major axis indeed lies along the $$x$$-direction. Using the eccentricity formula, we obtain

$$e_1=\sqrt{1-\frac{b_1^2}{a_1^2}}=\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}.$$

For the second ellipse $$E_2$$ we are given

$$\frac{x^2}{16}+\frac{y^2}{b^2}=1.$$

Here

$$a_2^2=16,\qquad b_2^2=b^2,$$

with $$16>b^2$$ so that the major axis is again along the $$x$$-direction. Applying the same eccentricity formula gives

$$e_2=\sqrt{1-\frac{b_2^2}{a_2^2}}=\sqrt{1-\frac{b^2}{16}}.$$

It is stated that the product of the two eccentricities equals $$\dfrac12$$. Hence

$$e_1\,e_2=\frac{1}{\sqrt{3}}\;\sqrt{1-\frac{b^2}{16}}=\frac12.$$

To remove the square root, we square both sides:

$$\left(\frac{1}{\sqrt{3}}\right)^2\!\left(1-\frac{b^2}{16}\right)=\left(\frac12\right)^2.$$

Simplifying each factor, we have

$$\frac13\left(1-\frac{b^2}{16}\right)=\frac14.$$

Now we isolate the bracketed term by multiplying through by $$3$$:

$$1-\frac{b^2}{16}=\frac34.$$

Next, we move the fraction involving $$b^2$$ to the right side:

$$-\frac{b^2}{16}=\frac34-1=-\frac14.$$

Multiplying by $$-1$$ gives

$$\frac{b^2}{16}=\frac14.$$

Finally, multiplying both sides by $$16$$ yields

$$b^2=16\left(\frac14\right)=4.$$

The semi-minor axis of $$E_2$$ is therefore

$$b=\sqrt{b^2}=\sqrt{4}=2.$$

The (full) minor axis length is twice the semi-minor axis, so

$$\text{minor axis length}=2b=2\times2=4.$$

Hence, the correct answer is Option C.