Aren't they equal?

Quick 'n' dirty

Cut the sphere in top half and bottom half. On the cutting surface, the pressure in both directions is $P = \rho g R$ , and the resulting vertical force is $F_{in} = PA = \pi \rho g R^3$ on each half sphere.

Call the total force on the outside of each half sphere $F_{out}$ , then the buoyant force $F_B = \pm(F_{out} - F_{in})$ is equal to the weight of the displaced fluid. Since the displaced fluid has the shape of a half-sphere, we know that $F_B = mg = \rho V g = \tfrac23\pi \rho g r^3$ .

Thus $F_{out} = F_{in} \mp F_B = \pi \rho g R^3 \mp \tfrac23\pi\rho g R^3 = \begin{cases}\tfrac13\pi\rho gR^3 & \text{top} \\ \tfrac53\pi\rho gR^3 & \text{bottom} \end{cases}.$

The ratio is obviously $\frac{\tfrac 13\pi\rho gR^3 }{\tfrac 53\pi\rho gR^3} = \frac 1 5 = \boxed{0.2}.$

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Standard approach

Choose units in which the radius is 1, and $\rho g = 1$ .

Consider a surface element $dA = \cos\theta\:d\theta\:d\phi,$ where $\theta$ is the "latitude" and $\phi$ the "longitude".

On this element acts a pressure $P = \rho g y = 1 - \sin\theta$ . Since all horizontal force components cancel, we only consider the vertical force component, $dF = P\sin\theta\:dA = (1 - \sin\theta)\sin \theta\cos\theta\:d\theta\:d\phi.$ Integrate this over the half-sphere: $F = \int_0^{\pm \pi/2} \int_0^{2\pi} d\phi\:d\theta\:(1 - \sin\theta)\sin\theta\cos\theta = 2\pi \int_0^{\pm \pi/2} d\theta (\sin\theta\cos\theta - \sin^2\theta\cos\theta).$ $= 2\pi \left(\int_0^{\pm \pi/2} \tfrac12 \sin 2\theta\:d\theta - \int_0^{\pm\pi/2} \sin^2\theta\:d\sin\theta\right) = 2\pi\left(\left[-\tfrac14\cos2\theta\right]_0^{\pm \pi/2} - \left[\tfrac13\sin^3\theta\right]_0^{\pm\pi/2}\right)$ $= 2\pi\left(\tfrac12 \mp \tfrac13\right) = \tfrac13\pi\ (\text{top}),\ \tfrac53\pi\ (\text{bottom}).$ Thus the ratio is $\frac{\tfrac 13\pi}{\tfrac 53\pi} = \frac 1 5 = \boxed{0.2}.$

Hydrostatic force is the weight of water above, the volume of sphere inscribed in a cylinder is 2/3 the volume of cylinder.Above the sphere is 1/6 and below the sphere is 1/6....hence the net ratio is (1/6)/(5/6)=0.2.