To be able to answer this question I extract 2N2222 BJT's model from Multisim.
ro ≈ (11.3V + 10V)/3.73mA ≈ 5.7kΩ
Therefore the voltage gain will be around
Av ≈ (Rc||RL||ro)/re ≈ (5kΩ||5.7kΩ||100kΩ)/6.9Ω ≈ 2.6kΩ/6.9Ω ≈ 376V/V
And if we do this, we see that Early's voltage (VAF) has a very low-value VAF = 10V. Thus, the transistor output resistance (ro) will be extremely low..MODEL 2N2222_Multisim npn
+IS=1.87573e-15 BF=153.575 NF=0.897646 VAF=10
+IKF=0.410821 ISE=3.0484e-09 NE=4 BR=0.1
+NR=1.00903 VAR=1.92063 IKR=4.10821 ISC=1.94183e-12
+NC=3.92423 RB=8.70248 IRB=0.1 RBM=0.1
+RE=0.111394 RC=0.556972 XTB=1.76761 XTI=1
+EG=1.05 CJE=1.67272e-11 VJE=0.83191 MJE=0.23
+TF=3.573e-10 XTF=0.941617 VTF=9.22508 ITF=0.0107017
+CJC=9.98785e-12 VJC=0.760687 MJC=0.345235 XCJC=0.9
+FC=0.49264 CJS=0 VJS=0.75 MJS=0.5
+TR=3.55487e-06 PTF=0 KF=0 AF=1
ro ≈ (11.3V + 10V)/3.73mA ≈ 5.7kΩ
Therefore the voltage gain will be around
Av ≈ (Rc||RL||ro)/re ≈ (5kΩ||5.7kΩ||100kΩ)/6.9Ω ≈ 2.6kΩ/6.9Ω ≈ 376V/V