Voltage drop on each diode - connected in series
I pretty much agree with your approach. Setting \$I_\text{CAL}=1\:\text{mA}\$ and \$V_\text{CAL}=700\:\text{mV}\$ and given \$\eta=1\$, the complete equation (using only Shockley diode equation) I get it, is:

$$R=\frac{10\:\text{V}-3\:\text{V}}{I_\text{CAL}\cdot\frac{e^{V_D/V_T}-1}{e^{V_\text{CAL}/V_T}-1}}$$

(Noting that \$V_D=750\:\text{mV}\$.)

Discounting the \$-1\$ terms, this simplifies a little:

$$\begin{align*}R&amp;\approx\frac{7\:\text{V}}{I_\text{CAL}\cdot e^{{\left(V_D-V_\text{CAL}\right)}/V_T}}\\\\&amp;\approx \frac{7\:\text{V}}{1\:\text{mA}\cdot e^{{\left(750\:\text{mV}-700\:\text{mV}\right)}/V_T}}\\\\&amp;\approx 7\:\text{k}\Omega \:\cdot e^{{-50\:\text{mV}}/V_T}\end{align*}$$

Depending on the room temperature value for \$V_T\$ that you use (Spice defaults to about \$25.876\:\text{mV}\$), you will roughly get \$950\:\Omega\lt R\lt 1020\:\Omega\$.

I've rounded your value, which I take to be at the low end of this range. But I think your approach is solid.