Whether Internal source resistance will have effect on the circuit
From the given input resistance \$r_{i}=20\;Ohms\$ I deduce that the transconductance is \$g_{m}=\frac{1}{r_{i}}=0.05A/V\$. Because of

$$
I_{c}=g_{m} \times V_{t} \quad(V_{t}=temeperature\: voltage, V_{t}=25mV)
$$

we can calculate the DC current \$I_{c}=I_{e}=0.05 \times 25 mA=1.25 mA\$ (neglecting \$I_{b}\$).

The voltage gain referred to the base node is \$A_{1}=g_{m} \times (r_{o}||R_{L})\$ and the gain referred to the signal source (with internal \$R_{s}=100\;ohms\$) is \$A_{0}=A_{1} \times r_{i}/(r_{i}+R_{s})\$.

For an input signal of Vi=0.5 V the resulting signal ouput is app. 4V.