Since $\beta$ is mentioned use re-model. First carry out dc analysis for Ie1 and Ie2, so all connected capacitors acts as open circuit. Hence circuit becomes,

As $\beta1=\beta2=220$, then Ie1 will be equal to Ie2

By Voltage divider rule,

$V_{R2}=\frac{R2}{R1+R2}Vcc$

$V_{R2}=\frac{2.2K}{22K+2.2K}\times15$

$V_{R2}=1.36V$

By KVL,

$V_{R2}-V_{BE}=V_{E1}$

$1.36V-0.7V=V_{E1}$

Let,$V_{BE}$=0.7V

$V_{E1}=0.663V$

$Ie1=\frac{V_{E1}}{Re1}=\frac{0.663V}{150\Omega}$

Ie1=4.424mA …