voltage analysis of BJT
It would be nice if you'd use the schematic editor that's available here, when you are forming (or editing) your question. I'll use it right now:







simulate this circuit &ndash; Schematic created using CircuitLab

It's pretty clear that the BJT is saturated, as shown on the right. From the right diagram, we have this nodal analysis:

$$\frac{V_E}{R_E}+\frac{V_E+V_{CE}}{R_C}=\frac{V_{CC}}{R_C}+1\:\textrm{mA}$$

The solution of which is:

$$V_E=\left(V_{CC}-V_{CE} + R_C\cdot 1\:\textrm{mA}\right)\cdot\frac{R_E}{R_E+R_C}$$

We also know that \$V_B=V_E+V_{BE}\$. One might quibble about the exact value of \$V_{CE}\$ and \$V_{BE}\$, but in this case it's not worth quibbling over. For now, set \$V_{CE}=0\:\textrm{V}\$ and \$V_{BE}=700\:\textrm{mV}\$. (You can always apply different values, later.) This works out to:

$$\begin{align*}
V_E&amp;=\left(12\:\textrm{V}-0\:\textrm{V} + 4.7\:\textrm{k}\Omega\cdot 1\:\textrm{mA}\right)\cdot\frac{1\:\textrm{k}\Omega}{1\:\textrm{k}\Omega+4.7\:\textrm{k}\Omega}\\\\
&amp;\approx 2.93\:\textrm{V}\\\\
V_B&amp;=V_E+V_{BE}\\\\
&amp;\approx 3.63\:\textrm{V}
\end{align*}$$

From that, it's clear that:

$$V_{R_B}=V_{CC}-V_B=12\:\textrm{V}-3.63\:\textrm{V}\approx 8.37\:\textrm{V}$$

Which isn't the answer you say it should be.

As a double-check, let's compute a few things:

$$\begin{align*}
I_C&amp;=\frac{V_{CC}-V_E-V_{CE}}{R_C}\\\\
&amp;=\frac{12\:\textrm{V}-2.93\:\textrm{V}-0\:\textrm{V}}{4.7\:\textrm{k}\Omega}\\\\
&amp;\approx 1.93\:\textrm{mA}\\\\
I_E&amp;=I_B+I_C\\\\&amp; \approx 2.93\:\textrm{mA}\\\\
V_E&amp;=R_E\cdot I_E\\\\
&amp;=1\:\textrm{k}\Omega\cdot 2.93\:\textrm{mA}\\\\
&amp;\approx 2.93\:\textrm{V}
\end{align*}$$

That last value confirms the original calculation made at the outset, which demonstrates that it all works out.



You can play around with the inputs of \$V_{CE}\$ and \$V_{BE}\$ and I think you'll see it doesn't really change that much. Either you didn't provide full and complete and accurate information here, or else there is a problem with the answer you mentioned in one of your comments here.