Need solution for this CB BJT circuit
As always, it's helpful to first draw the DC and AC circuits.

DC circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

The operating point is evident by inspection:

$$I_C = \frac{\beta}{1 + \beta}I_2  = \alpha I_2 $$

$$V_C = I_C(\frac{75\Omega}{\alpha} + \frac{100k\Omega}{\beta}) + V_{BE} $$



Update to address comment:


  I can't perfectly grasp your equation for Vcc.I think understand you
  divide resistance with beta and alpha to make them equivalent
  resistance looking through C.


Assuming you meant \$V_C\$ rather than \$V_{CC}\$, by KVL we have

$$V_C = V_E + V_{BE} + V_{R1}$$

We have

$$V_E = I_E R_S = \frac{I_C}{\alpha}R_S $$

and

$$V_{R1} = I_B R_1 = \frac{I_C}{\beta}R_1$$

Thus

$$V_C = I_C(\frac{R_S}{\alpha} + \frac{R_1}{\beta}) + V_{BE} $$



AC circuit:







simulate this circuit

The small-signal circuit is thus







simulate this circuit

This is a straightforward circuit to solve.  What have you tried so far?