Step-by-step Solution:

1. Analyze the given ellipse E₁ and the rectangle R. The equation of the first ellipse is ${E_1}: {{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$. This is a standard ellipse centered at the origin $(0,0)$. The semi-major axis is $a = \sqrt{9} = 3$ (along the x-axis). The semi-minor axis is $b = \sqrt{4} = 2$ (along the y-axis).

   The ellipse $E_1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. This means the sides of the rectangle are tangent to the ellipse at its vertices. The vertical sides are the lines $x = \pm 3$. The horizontal sides are the lines $y = \pm 2$. Therefore, the vertices of the rectangle $R$ are $(\pm 3, \pm 2)$. Specifically, the four vertices are $(3, 2)$, $(-3, 2)$, $(-3, -2)$, and $(3, -2)$.

2. Determine the equation of the second ellipse E₂. The second ellipse $E_2$ circumscribes the rectangle $R$. This means $E_2$ passes through all four vertices of $R$. Since the rectangle is symmetric with respect to both coordinate axes, the circumscribing ellipse $E_2$ must also be centered at the origin. Let the equation of $E_2$ be: ${{{x^2}} \over {A^2}} + {{{y^2}} \over {B^2}} = 1$

   Since $E_2$ passes through the vertices of $R$, we can pick one vertex, say $(3, 2)$, and substitute its coordinates into the equation of $E_2$: ${{{3^2}} \over {A^2}} + {{{2^2}} \over {B^2}} = 1$ {{9 \over {A^2}} + {{4 \over {B^2}} = 1} \quad \cdots (1)

   We are also given that the ellipse $E_2$ passes through the point $(0, 4)$. Substituting these coordinates into the equation of $E_2$: ${{{0^2}} \over {A^2}} + {{{4^2}} \over {B^2}} = 1$ ${{16 \over {B^2}} = 1} \implies {B^2 = 16}$

3. Solve for the parameters of E₂. Now we substitute the value of $B^2 = 16$ back into equation (1): {{9 \over {A^2}} + {{4 \over {16}} = 1} {{9 \over {A^2}} + {{1 \over 4} = 1} ${{9 \over {A^2}} = 1 - {1 \over 4} = {3 \over 4}}$ ${A^2 = 9 \times {4 \over 3} = 12}$

   So, the equation of the ellipse $E_2$ is: ${{{x^2}} \over {12}} + {{{y^2}} \over {16}} = 1$

4. Calculate the eccentricity of the ellipse E₂. For the ellipse $E_2$, we have $A^2 = 12$ and $B^2 = 16$. Since $B^2 > A^2$, the major axis of the ellipse is along the y-axis. The semi-major axis is $b' = \sqrt{16} = 4$. The semi-minor axis is $a' = \sqrt{12} = 2\sqrt{3}$.

   The eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{{\text{(semi-minor axis)}^2}}{{(\text{semi-major axis)}^2}}}$. In our case, this is $e = \sqrt{1 - \frac{A^2}{B^2}}$. $e = \sqrt{1 - {{12} \over {16}}} = \sqrt{1 - {3 \over 4}} = \sqrt{{1 \over 4}}$ $e = {1 \over 2}$

   The eccentricity of the ellipse $E_2$ is $1/2$.

5. Conclusion. Comparing our result with the given options: A: ${{\sqrt 2 } \over 2} \approx 0.707$ B: ${{\sqrt 3 } \over 2} \approx 0.866$ C: ${{1 \over 2}} = 0.5$ D: ${{3 \over 4}} = 0.75$

   Our calculated eccentricity matches option C.