A vertical pole stands at the center of a regular hexagonal field....

Let the side length of the regular hexagon be $a$ and the height of the vertical pole be $h$.

The distance from the center of a regular hexagon to any of its vertices is equal to its side length $a$.
Given that the angle of elevation from a vertex is $\alpha$, we have:
$\tan \alpha = \dfrac{h}{a} = 2 \implies h = 2a$

The distance from the center of the hexagon to the midpoint of any side is the apothem, which is $a\cos(30^{\circ}) = \dfrac{a\sqrt{3}}{2}$.
The angle of elevation from the midpoint of a side is $\beta$, so:
$\tan \beta = \dfrac{h}{\dfrac{a\sqrt{3}}{2}} = \dfrac{2h}{a\sqrt{3}}$

Substitute $h = 2a$ into the equation:
$\tan \beta = \dfrac{2(2a)}{a\sqrt{3}} = \dfrac{4}{\sqrt{3}}$

We need to find the value of $3 \sec^2 \beta$:
$3 \sec^2 \beta = 3(1 + \tan^2 \beta)$
$3 \sec^2 \beta = 3\left(1 + \left(\dfrac{4}{\sqrt{3}}\right)^2\right)$
$3 \sec^2 \beta = 3\left(1 + \dfrac{16}{3}\right) = 3 + 16 = 19$

Answer: $19$