= Solution
Let the approximately conserved <magnetic flux> be $\Phi$. Then
$$
\Phi=\mu_0H_{\rm gap}\pi r^2.
$$
The length $L$ shown in the original diagram is the untapered core arc; the two tapers are separate lengths $s$ each. In the untapered core, <flux> conservation gives
$$
H_R=\frac{\Phi}{\mu_0\mu\pi R^2}
=\frac{H_{\rm gap}}\mu\frac{r^2}{R^2}.
$$
For one taper use distance $u$ from its narrow end. The straight taper in the diagram has local radius
$$
a(u)=r+\frac{R-r}{s}u,\qquad0\le u\le s,
$$
and <magnetic field intensity>
$$
H(u)=\frac{\Phi}{\mu_0\mu\pi a(u)^2}
=\frac{H_{\rm gap}}\mu\frac{r^2}{a(u)^2}.
$$
Its contribution to the <line integral> in the <magnetostatic Ampère law> is found explicitly:
$$
\int_0^s\frac{du}{a(u)^2}
=\frac{s}{R-r}\left(\frac1r-\frac1R\right)=\frac{s}{rR},
\qquad
\int_0^sH(u)\,du=\frac{H_{\rm gap}}\mu\frac{sr}{R}.
$$
Adding the gap, the untapered arc and both tapers gives
$$
NI=H_{\rm gap}h+\frac{H_{\rm gap}}\mu\frac{r^2}{R^2}L
+2\frac{H_{\rm gap}}\mu\frac{sr}{R}.
$$
Therefore
$$
\boxed{H_{\rm gap}=\frac{\mu NI}{\mu h+L(r/R)^2+2s(r/R)}.}
$$
This uses a linear radius profile, approximately uniform <magnetic flux density> over each section, a linear constant <magnetic permeability>, and negligible leakage and <magnetic fringing>. For a non-linear taper profile $a(u)$ the honest generalization is
$$
H_{\rm gap}=\frac{\mu NI}{\mu h+Lr^2/R^2+2r^2\int_0^sdu/a(u)^2}.
$$
Thus an arbitrary smooth taper with the same end radii and length need not give the printed formula; the straight-sided geometry supplies the additional information.
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