A Note About A Certain Simplifiation
When we say that the field between these inductors is homogeneous, it's actually a simplification. These inductors have an effectively zero magnetic field outside of them because the field produced from one segment of wire effectively cancels out the field produced from a segment on the other side. This simplification lets us say: outside of an inductor, there is no magnetic field from that inductor.
This simplification breaks down as you get close to the solenoid/inductor. To be more specific, when the distance from one side of the inductor to the other diverges from zero, this simplification no longer holds. In practice, you rarely do anything where this simplification doesn't apply.
The Answer
Given the simplification that the magnetic field produced from a solenoid is zero when outside the coils of that solenoid, you can say that there is no magnetic field from the interior inductor in that space. Therefore it is as if that interior solenoid is not there. Assigning the variables to the larger, exterior solenoid with a subscript 1, you get $B = B_{1_{in}} = \frac{\mu\mu_0}l{I_1n_1}$ when $R_2<r<R_1$.
If we label the pds across successive vertical resistors "$V_n$", "$V_{n+1}$" and so on we can write the Kirchoff current law at a junction as$$\frac{V_n-V_{n+1}}{R_A}=\frac{V_{n+1}}{R_B}+\frac{V_{n+1}-V_{n+2}}{R_A}$$
in which $R_A$ and $R_B$ are the values of a horizontal and vertical resistor respectively. If you're happy with the assumption that $\frac{V_{n+2}}{V_{n+1}}=\frac{V_{n+1}}{V_{n}}\ [=\alpha\ \,\text{say}]\ $ throughout the ladder, then you have a quadratic equation for $\alpha$. Knowing $\alpha$ you can find the input current for a given input pd, and hence the input resistance.
Possibly rather less convincing than the 'no change with an extra section' method, but at least it's rather different! And it does give the same answer!
A few numerical investigations… Suppose the right hand end of the ladder is a vertical resistor (2 ohm) and the last section of the ladder is a horizontal resistor (1 ohm) connected to that vertical resistor. Then the resistance, $R_1$, of the last section seen from the left is 3 ohm. The resistance, $R_2$, of the last two sections together is, seen from the left, 11/5 ohm, then we have 43/21 ohm, 171/85 ohm. These resistances, I found, fit the pattern$$R_{m}\text{\ohm}=2+\frac{3}{4^{m}-1}.$$Clearly the convergence to 2 ohm will be very rapid!
Addition to post
Simplify notation by calling the horizontal resistance "X" and the vertical resistance "$\beta X$". Call the resistance of a ladder of m sections (one horizontal resistor joined on the right to a vertical resistor down to a 'ground' rail), "$r_{m}X$". Then, assuming that adding one more section on to the left of an infinite ladder will make no difference to its resistance we have$$r_{\infty}X=X+\frac{\beta X.r_{\infty}X}{\beta X+r_{\infty}X}\ \ \ \text{giving}\ \ \ \ \ r_{\infty}=\frac{1}{2}\left(1+\sqrt{1+4\beta}\right)$$
$\beta=2$ gives $r_{\infty}=2$, as we know. Certain other values of $\beta$ will also give rational values of $r_{\infty}$, for example, $\beta \ =$12, 6, 2, 3/4, 5/16, 9/64 give respectively, $r_{\infty}=\ $4, 3, 2, 3/2, 5/4, 9/8. No doubt in each of these cases we can find a general formula for $r_m$ as we did for when $\beta=2$, by inspecting the bit left over when we subtract $r_\infty$ from $r_1$, $r_2$ and so on.
For most values of $\beta$, this procedure won't stand a chance of working, because $r_{\infty}$ will be irrational, while $r_1$, $r_2$ and so on are clearly rational if $\beta$ is rational! Instead, for the general case, we use binomial series... Expanding the expression for $r_\infty$ given above we get
$$r_\infty=1+\beta-\beta^2+2\beta^3-5\beta^4+14\beta^5-42\beta^6+132\beta^7-429\beta^8+...$$
But clearly, $r_1=1+\beta$.
So $r_1$ agrees with $r_\infty$ up to the second term!
We find without much difficulty that$$r_2=1+\frac{\beta(1+\beta)}{1+2\beta}$$
Expanding$\frac{1}{1+2\beta}$ binomially and multiplying out, we get$$r_2=1+\beta-\beta^2 +2\beta^3 -4\beta^4+...$$
So we now have agreement with $r_\infty$ up to the fourth term!
Continuing…$$r_3=1+\frac{\beta(1+3\beta +\beta^2)}{1+4\beta +3\beta^2}$$and this gives us$$r_3=1+\beta-\beta^2+2\beta^3-5\beta^4+14\beta^5-41\beta^6+…$$
So we now have agreement with $r_\infty$ up to the sixth term!
This may not have been very elegant, but I'm certainly now convinced! [Presumably $\beta$ is restricted to being less than 1 for this method to be valid.] Thank you for this interesting problem.
Best Answer
You can start by choosing an origin, and then define $r_1$, $r_2$, and $r_3$ in terms of of the $x$ and $y$ coordinate of this system. Then, realize that $\vec{B}$ has two components, which should both be zero for the magnetic field to vanish. You get two equations (one per component) in two unknowns (the coordinates), which you should be able to solve. You'll get two solution, one on either side of the horizontal axis.