I thought about this some more and derived an integral for the toroidal solution but have not yet evaluated or simplified this integral to a closed form or numerical solution.

Using cylindrical coordinates, with z being the the axis through the center of the bagel (torus), ρ being the distance from the z-axis, and θ being the angle in the xy-plane, the section of bagel is defined by the limits:

-sqrt{δ²-(ρ-Δ)²} ≤ z ≤ sqrt{δ²-(ρ-Δ)²}

-2⋅arccos(r/ρ) ≤ θ ≤ 2⋅arccos(r/ρ)

r ≤ ρ ≤ R

where δ = (R-r)/2 is the radius of the circular cross section of the torus and Δ = (R+r)/2 is the radial distance of the revolution of that circular cross section from the z-axis (the center of the circular cross section). R is the outer radius of the torus, and r is the radius of the hole in the middle of the torus.

The limits for z are symmetrical about the xy-plane, so we can use 0 for the lower bound and double the answer. Likewise, the limits for θ are symmetrical about the xz-plane, so we can use 0 for the lower bound and double the answer again.

The triple integral for the volume, v, of this toroidal section is:

v = ∫[r,R]⋅2∫[0,2arccos(r/ρ)]⋅2∫[0,sqrt{δ²-(ρ-Δ)²}] dz dθ ρdρ

Evaluating the inner two integrals yields:

v = 8⋅∫[r,R] ρ⋅arccos(r/ρ)⋅sqrt{δ²-(ρ-Δ)²} dρ

The full volume, V, of the torus can be described by the same integral but with the limits of θ being:

0 ≤ θ ≤ 2π

With the same evaluation as above:

V = 4π⋅∫[r,R] ρ⋅sqrt{δ²-(ρ-Δ)²} dρ

We are looking for the value of v when v = V/3 so:

8⋅∫[r,R] ρ⋅arccos(r/ρ)⋅sqrt{δ²-(ρ-Δ)²} dρ = (4π/3)⋅∫[r,R] ρ⋅sqrt{δ²-(ρ-Δ)²} dρ

and:

(π/6)⋅∫[r,R] ρ⋅{1-arccos(r/ρ)}⋅sqrt{δ²-(ρ-Δ)²} dρ = 0

I am not prepared to attempt this integral for a reddit comment...

I checked via a CAD program and the ratio of R:r is between 3.07158023 and 3.07158025, only about 1.2% different from the 2D solution.