How to calcuate base resistance from BJT circuit?
I think the following includes all of the elements you were asked to achieve, including make adjustments on \$V_{BE}\$ because of the higher collector current than the nominal value given. It's implied by the question, I suspect, that you need to adjust the \$700\:\textrm{mV}\$ figure.

$$\begin{align*}
I_C&amp;=\frac{10\:\textrm{V}-5\:\textrm{V}}{1\:\textrm{k}\Omega}\\\\
&amp;= 5\:\textrm{mA}\\\\
I_B&amp;= \frac{I_C}{\beta}\\\\
&amp;= 50\:\mu\textrm{A}\\\\
V_{BE}&amp;\approx 700\:\textrm{mV} + 26\:\textrm{mV}\cdot\operatorname{ln}\left(\frac{5\:\textrm{mA}}{1\:\textrm{mA}}\right)\\\\
&amp;\approx 742\:\textrm{mV}\\\\
R_B&amp;=\frac{10\:\textrm{V}-V_{BE}}{I_B}\\\\
&amp;\approx 185.2\:\textrm{k}\Omega
\end{align*}$$

However, this doesn't agree with the answer you said is right. But there it is, anyway. By the way, \$\frac{k T}{q}\approx 26\:\textrm{mV}\$ at room temperature. That's where that number came from, above.