BJT Problem to find emitter current
Since you haven't selected your own answer yet, here's mine that may reflect your understanding:







simulate this circuit &ndash; Schematic created using CircuitLab

All I've done in this transition is to set up the Thevenin \$V_\text{TH}\$ and \$R_\text{TH}\$ from the original biasing pair, \$R_1\$ and \$R_2\$. This makes it very easy to perform the following KVL:

$$V_\text{CC}-I_\text{E}\cdot R_E-V_\text{BE}-I_\text{B}\cdot R_\text{TH}=V_\text{TH}$$

Some very simple algebraic manipulation results in:

$$I_\text{E}=\frac{V_\text{CC}-V_\text{TH}-V_\text{BE}}{\frac{R_\text{TH}}{\beta+1}+R_E}$$

Plugging in the values of \$V_\text{TH}=3\:\text{V}\$, \$R_\text{TH}=2.4\:\text{k}\Omega\$, \$V_\text{CC}=15\:\text{V}\$, \$\beta=49\$, \$V_\text{BE}=700\:\text{mV}\$, and \$R_\text{E}=3\:\text{k}\:\Omega\$, we find that \$I_\text{E}\approx 3.71\:\text{mA}\$.