Solution
= Solution
At each height, the removed <circular sector> has the same <angle> and $4/5$ of the original <radius>. Its <area> and hence its <volume> are therefore multiplied by $(4/5)^2=16/25$. Equating the removed <volume> to that of the reformed solid <sphere> gives
$$
\frac43\pi a^3=\frac{16}{25}V=\frac{2\pi R^2H}{25}(3-e^{-2}).
$$
Consequently
$$
\boxed{a=\left[\frac{3R^2H}{50}(3-e^{-2})\right]^{1/3}.}
$$