Calculating Ratio of Stability Constants
You used all the information from the question except the line that would have helped you solve it


  The concentrations of iron-containing species satisfy the relations $[\ce{Fe^2+}] = [\ce{Fe(CN)_6^4-}]$ and $[\ce{Fe^3+}] = [\ce{Fe(CN)_6^3-}]$


From your first cell you find that

$$\frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} = 103.97$$

which means that in your second cell

$$\frac{[\ce{Fe(CN)6^4-}]}{[\ce{Fe(CN)6^3-}]} = 103.97$$

and $E^\circ$ for the cell with cyanide is $+0.361~\mathrm{V}$. (You have a typo in your equation, the 4- should be in the numerator and 3- should be in the denominator. Also your equation is missing units but I won't bother with that.) This is quite far off from the accepted literature value (around $+0.430~\mathrm{V}$) but let's move on.


  Now I know that the ratio is just the sum of the 2 logs in the Nernst equations that I have wrote above


No, that is wrong. The ratio of the stability constants refers to that expression you have written, when all of the compounds are placed into the same solution. You cannot simply use the two equations earlier because those describe two separate solutions.

You have to use the difference in $E^\circ$ to work out the ratio of $\beta$.

The way to do this would be to realise that if you take the difference between the $E^\circ$ you get

$$\begin{align}
\ce{Fe^3+ + Fe(CN)6^4-} &amp;\longrightarrow \ce{Fe^2+ + Fe(CN)6^3-} &amp; E^\circ &amp;= +0.410~\mathrm{V}
\end{align}$$

The equilibrium constant for this reaction is the quantity that you are looking for

$$K = \frac{[\ce{Fe^2+}][\ce{Fe(CN)6^3-}]}{[\ce{Fe^3+}][\ce{Fe(CN)6^4-}]}$$

and so

$$K = \exp\left(\frac{-\Delta_\mathrm{r} G^\circ}{RT}\right) = \exp\left(\frac{-nFE^\circ}{RT}\right) = 8.589 \times 10^6$$