Alright, I've come up with a proof in what I think is the right flavor.
Take a sphere with radius $r$, and consider the upper hemisphere. For each $n$, we will construct a solid out of stacks of pyramidal frustums with regular $n$-gon bases. The stack will be formed by placing $n$ of the $n$-gons perpendicular to the vertical axis of symmetry of the sphere, centered on this axis, inscribed in the appropriate circular slice of the sphere, at the heights $\frac{0}{n}r, \frac{1}{n}r, \ldots,\frac{n-1}{n}r $ . Fixing some $n$, we denote by $r_\ell$ the radius of the circle which the regular $n$-gon is inscribed in at height $\frac{\ell}{n}r$ . Geometric considerations yield $r_\ell = \frac{r}{n}\sqrt{n^2-\ell^2}$ .
As noted in the question, the area of this polygonal base will be $\frac{n}{2}r_\ell^2 \sin\frac{2\pi}{n}$ for each $\ell$ . I am not sure why (formally speaking) it is reasonable to assume, but it appears visually (and appealing to the 2D case) that the sum of the volumes of these frustums should approach the volume of the hemisphere.
So, for each $\ell = 1,2,\ldots,n-1$, the term $V_\ell$ we seek is $\frac{1}{3}B_1 h_1 - \frac{1}{3}B_2 h_2 $, the volume of some pyramid minus its top. Using similarity of triangles and everything introduced above, we can deduce that
$$
B_1 = \frac{n}{2}r_{\ell-1}^2 \sin\frac{2\pi}{n}~,~B_2 = \frac{n}{2}r_\ell^2 \sin\frac{2\pi}{n} ~,~h_1 = \frac{r}{n}\frac{r_{\ell-1}}{r_{\ell-1}-r_{\ell}}~,~h_2=\frac{r}{n}\frac{r_{\ell}}{r_{\ell-1}-r_{\ell}} ~~.
$$
So, our expression for $V_\ell$ is
$$
\frac{r}{6} \sin\frac{2\pi}{n} \left\{ \frac{r_{\ell-1}^3}{r_{\ell-1}-r_{\ell}} - \frac{r_{\ell}^3}{r_{\ell-1}-r_{\ell}} \right\} = \frac{\pi r}{3n} \frac{\sin\frac{2\pi}{n}}{2\pi/n} \left\{ r_{\ell-1}^2 + r_\ell^2 + r_{\ell-1}r_\ell \right\}
$$ $$
= \frac{\pi r^3}{3n^3} \frac{\sin\frac{2\pi}{n}}{2\pi/n} \left\{ (n^2 - (\ell-1)^2) + (n^2-\ell^2) + \sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)} \right\} ~~.
$$
So, we consider $ \lim\limits_{n\to\infty} \sum_{\ell=1}^{n-1} V_\ell$ . The second factor involving sine goes to 1, and we notice that each of the three terms in the sum is quadratic in $\ell$, and so the sum over them should intuitively have magnitude $n^3$. Hence, we pass the $\frac{1}{n^3}$ into the sum and evaluate each sum and limit individually, obtaining 2/3, 2/3, and 2/3 respectively (the first two are straightforward, while the third comes from the analysis in this answer).
Thus, we arrive at $\frac{\pi r^3}{3} (2/3+2/3+2/3) = \frac{2}{3}\pi r^3$ as the volume of a hemisphere, as desired.
So was this too excessive or perhaps worth it? I'll leave that to all of you. :)
You should substitute for $h$ to get:
$$V=\dfrac{1}{3}\dfrac{b\sqrt{k^2-2kb^2}}{2}$$
Then differentiate w.r.t to $b$:
$$\dfrac{dV}{db}=\dfrac{1}{6}\left(\dfrac{1}{2\sqrt{b^2k^2-2kb^4}}\times(2bk^2-8kb^3)\right)=\dfrac{1}{6}\left(\dfrac{bk}{\sqrt{b^2k^2-2kb^4}}\times(k-4b^2)\right)=0\\
\implies b=\dfrac{\sqrt k}{2}$$
Of course, another way to do this is by trial and error, which is not very effective. The above method is the most effective, and works very well.
Best Answer
I don't think that the answer provided by Anthony is correct. The problem is that the slant length s is not constant, I found here https://rechneronline.de/pi/elliptic-cone.php an approximating formula for the lateral area of an elliptic cone (not truncated!) $$ A := \frac{1}{2} \pi ( a \sqrt{ b^2 + h^2} + b \sqrt{a^2 + h^2 }) $$ i think it can be further generalized for a truncated cone