What is the percent of potassium as K₂O in the fertilizer sample?
The solubility has been given solely to justify the application of the given method of gravimetric analysis.
Since the majority of potassium salts are well or moderately soluble in water, for this analysis it's assumed that entire potassium content is brought to the solution. 
Since the solubility value of potassium tetraphenylborate is over a million times smaller than the corresponding value for the analyte (sodium tetraphenylborate), it's safe to speak of the completeness of the analysis as all dissolved potassium species were isolated as tetraphenylborate salt and analyzed.

As for the mass difference, that shouldn't be of your concern as the content of the fertilizer isn't given in the first place and you cannot make any assumptions regarding molecular masses of its constituents.
Plus, tetraphenylborate $\ce{B(C6H5)4-}$ is a bulky anion of substantial molecular mass $(M(\ce{B(C6H5)4-}) = \pu{319.23 g mol-1})$, which is much higher in comparison to "usual suspects" found in fertilizers, such as phosphates $(M(\ce{PO4^3-}) = \pu{94.97 g mol-1})$ and nitrates $(M(\ce{NO3-}) = \pu{62.00 g mol-1})$.

Long story short, your answer seems correct to me, but I propose to solve the problem algebraically first, starting with what has to be determined and omit intermediate calculations, unless you were explicitly asked to perform any.
To illustrate my approach and use of the notations for physical quantities and units, I leave my solution here.

Content of potassium expressed via potassium oxide is a mass fraction $ω(\ce{K2O})$, which is the ratio between the mass of potassium oxide $m(\ce{K2O})$ and the total mass of fertilizer taken for analysis $m_\mathrm{tot}$:

$$ω(\ce{K2O}) = \frac{m(\ce{K2O})}{m_\mathrm{tot}}$$

Unknown mass $m(\ce{K2O})$ can be found from the corresponding amount $n(\ce{K2O})$ and molecular mass $M(\ce{K2O})$:

$$m(\ce{K2O}) = n(\ce{K2O})\cdot M(\ce{K2O})$$

In turn, amount $n(\ce{K2O})$ can be determined from the reaction's stoichiometry:

$$n(\ce{K2O}) = \frac{1}{2}n(\ce{K+}) = \frac{1}{2}n(\ce{KB(C6H5)}) = \frac{m(\ce{KB(C6H5)})}{2\cdot M(\ce{KB(C6H5)})}$$

Now, since we are at the point when no unknowns are left, lets put everything together so that the first equation can be rewritten as such:

$$
\begin{align}
ω(\ce{K2O}) &amp;= \frac{m(\ce{KB(C6H5)})\cdot M(\ce{K2O})}{2\cdot m_\mathrm{tot}\cdot M(\ce{KB(C6H5)})} \\
  &amp;= \frac{\pu{12.4393 g}\cdot\pu{94.20 g mol-1}}{2\cdot\pu{10.7875 g}\cdot\pu{358.32 g mol-1}} \\
  &amp;= \pu{0.152}~\text{or}~15.2\%
\end{align}
$$