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

As stage 1 and stage 2 are identical & value of$\beta1=\beta2$=220

Ie1=Ie2=4.424mA

AC equivalent of the circuit becomes,

Rin (Input Resistance):

$Rin=R1||R2||\beta re$

$re=\frac{26mV}{Ie1}=\frac{26mV}{4.424mA}$

$re=5.87\Omega$

Hence, Rin becomes

$Rin=22k||2.2k||220\times5.87$

$Rin = 0.784k\Omega$

Ro(Output Resistance):

$Ro=Rc2s=1k\Omega$

Av (Voltage Gain):

$Av2=\frac{Vo}{Vo1}=\frac{-Rc2}{re}$

$Av2=\frac{-1k\Omega}{5.87\Omega}$

Av2=-170.35

$Av1=\frac{-(Rc1||Rin2)}{re}$

But, $Rin2= R3||R4||\beta re =0.784k\Omega$

$Av1=\frac{(-1k\Omega||0.784k\Omega)}{5.87\Omega}$

Av1=-74.865

So Av is obtained by multiplying Av1 and Av2,

$Av=Av1 \times Av2$

$Av = (-170.35 \times -74.865)$

Av=12753.25