Hexagonal Pyramid.

Let $OP$ be the height $H$ of the hexagonal pyramid.

Let $E: (\dfrac{-x}{2}, \dfrac{\sqrt{3}}{2}, 0)$ , $A: (\dfrac{-x}{2}, \dfrac{-\sqrt{3}}{2}, 0)$ , $F: (-x,0,0)$ , $P: (0,0,H)$ .

$\vec{FP} = x\vec{i} + 0\vec{j} + H\vec{k} , \:\ \vec{EP} = \dfrac{x}{2}\vec{i} - \dfrac{\sqrt{3}}{2} x\vec{j} + H\vec{k}, \:\ \vec{AP} = \dfrac{x}{2}\vec{i} + \dfrac{\sqrt{3}}{2} x\vec{j} + H\vec{k}$

$\vec{u} = \vec{FP} \times \vec{EP} = \dfrac{\sqrt{3}}{2} xH \vec{i} -\dfrac{xH}{2}\vec{j} - \dfrac{\sqrt{3}}{2} x^2\vec{k}$ and $\vec{v} = \vec{FP} \times \vec{AP} = \dfrac{-\sqrt{3}}{2} xH \vec{i} -\dfrac{xH}{2}\vec{j} + \dfrac{\sqrt{3}}{2} x^2\vec{k}$

$\vec{u} \cdot \vec{v} = \dfrac{-x^2}{4} (2 H^2 + 3 x^2)$ and $|\vec{u}| = |\vec{v}| = \dfrac{x}{2} \sqrt{4H^2 + 3x^2} \implies \cos(\theta) = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \dfrac{-(2H^2 + 3x^2)}{4H^2 + 3x^2}$ , where $\theta$ is the angle between the two adjacent slant faces.

Let $V_{s} = \dfrac{4}{3}\pi R^3$ be the volume of the sphere.

The Area $A_{hexagon} = \dfrac{3\sqrt{3}}{2} x^2 \implies V_{p} = \dfrac{\sqrt{3}}{2} x^2 H$

In the right triangle above: $AC = H - R, \:\ BC = x, \:\ AB = R \implies$ $x^2 = 2HR - H^2 \implies V_{p}(H) = \dfrac{\sqrt{3}}{2} (2 H^2 R - H^3) \implies \dfrac{dV_{p}}{dH} = \dfrac{\sqrt{3}}{2} H(4R - 3H) = 0$ $H \neq 0 \implies H = \dfrac{4R}{3} \implies x^2 = \dfrac{8}{9}$

Using $H^2 = \dfrac{16 R^2}{9}, x^2 = \dfrac{8 R^2}{9}$ and $\cos(\theta) = \dfrac{-(2H^2 + 3x^2)}{4H^2 + 3x^2} \implies$ $\cos(\theta) = \dfrac{-7}{11} \implies \theta = 129.52^\circ$