Slant Hexagonal Pyramid.

Let $r$ be the length of a side of the hexagon and $h = GA$ be the height of the hexagonal pyramid.

$C:(\dfrac{r}{2},\dfrac{\sqrt{3}}{2},0), \:\ B:(-\dfrac{r}{2},\dfrac{\sqrt{3}}{2},0)$ and $G(r,0,h)$ .

$\vec{u} = -r\vec{i} + 0\vec{j} + 0\vec{k}$ and $\vec{v} = \dfrac{r}{2}\vec{i} - \dfrac{\sqrt{3}}{2}\vec{j} + h\vec{k} \implies \vec{u} X \vec{v} = 0\vec{i} + rh\vec{j} + \dfrac{\sqrt{3}}{2}r^2\vec{k} \implies$

$d = \dfrac{\sqrt{4h^2 + 3r^2}}{2} \implies A_{\triangle{GCB}} = A_{\triangle{GEF}} = \dfrac{1}{4}\sqrt{4h^2 + 3r^2}r$

Let $A = \dfrac{1}{4}\sqrt{4h^2 + 3r^2}r$

The volume $V = \dfrac{1}{4\sqrt{3}}r^2h = k \implies h = \dfrac{4\sqrt{3}}{r^2} \implies$ $A(r) = \dfrac{1}{4}\dfrac{\sqrt{192k^2 + 3r^6}}{r} \implies$ $\dfrac{dA}{dr} = \dfrac{3(r^6 - 32k^2)}{2r^2\sqrt{192k^2 +3r^6}} = 0 \implies r = (4\sqrt{2}k)^{\frac{1}{3}} \implies h = 4\sqrt{3}k(\dfrac{1}{4\sqrt{2}k})^{\frac{2}{3}}$

Using above diagram, Let $\theta = m\angle{GBQ} = m\angle{GFP} \implies$

$\cos(\theta) = \dfrac{|\vec{u} \cdot \vec{v}|}{|\vec{u}||\vec{v}|} = \dfrac{r}{2\sqrt{r^2 + h^2}} = \dfrac{1}{\sqrt{10}} \implies \theta \approx 71.565051^\circ$

$\implies \lambda = 180^\circ - \theta = \boxed{108.434949}$ .