Diode (Iterative analysis) with exponential model
Your basic equation for each diode's voltage, in this problem where you are told that when \$I_D=1\:\textrm{mA}\$ that \$V_D=700\:\textrm{mV}\$, should look like:

$$V_D= 700\:\textrm{mV} + n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)$$

This is easily read as two terms: the first term being the assumed voltage when the current actually is \$1\:\textrm{mA}\$ and the second term being the difference from that assumed voltage when the current itself is \$I_D\$. Note that if \$I_D=1\:\textrm{mA}\$, the logarithm will be zero so there will be no adjustment.

Now you are asked to find the value of the resistor if you want to achieve a stacked diode voltage of \$V_o=2.4\:\textrm{V}\$. Clearly, this means that the voltage across each diode (same current in each, all diodes assumed to use the exact same model) must be \$\frac{1}{3}\$rd as much, or \$V_D=800\:\textrm{mV}\$.

So, you are trying to solve for \$I_D\$ in the case where:

$$\begin{align*}
800\:\textrm{mV}&amp;= 700\:\textrm{mV} + n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)\\\\
800\:\textrm{mV}-700\:\textrm{mV} &amp;=  n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)\\\\
100\:\textrm{mV} &amp;=  n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)\\\\
\frac{100\:\textrm{mV}}{n\cdot V_T} &amp;= \operatorname{ln}\left(\frac{I_D}{1\:\textrm{mA}}\right)\\\\
e^\frac{100\:\textrm{mV}}{n\cdot V_T} &amp;= \frac{I_D}{1\:\textrm{mA}}\\\\
I_D&amp;=1\:\textrm{mA}\cdot e^\frac{100\:\textrm{mV}}{n\cdot V_T}
\end{align*}$$

For diodes with an emission coefficient of \$n=1\$ and where you are specifying that \$V_T=25\:\textrm{mV}\$ (I use the slightly higher \$V_T=26\:\textrm{mV}\$ as a rule), then I get \$I_D\approx 54.6\:\textrm{mA}\$. From this, I would compute: \$R= \frac{10\:\textrm{V}-2.4\:\textrm{V}}{54.6\:\textrm{mA}}\approx 139.2\:\Omega\$.

As you can see, I didn't use the 2.3 factor you included. But I did arrive at a very similar place. So, this probably just means I don't understand where you got that factor and how you applied it in the rest of your work.

Please note that I would write the following:

$$\begin{align*}
V_{D_1}&amp;= n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_1}}{I_S}\right)\\\\
V_{D_2}&amp;= n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_2}}{I_S}\right)\\\\
\therefore \Delta V_D=V_{D_2}-V_{D_1}&amp;=n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_2}}{I_S}\right)-n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_1}}{I_S}\right)\\\\
\Delta V_D&amp;=n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{D_2}}{I_{D_1}}\right)
\end{align*}$$

I still don't see where that 2.3 came from or how you applied it. But we got to similar places.

EDIT: Never mind. I think you are using LOG10, right? I think I understand better, now!