Question on DC analysis of BJT - Common Base
$$ I_E= \frac{(V_{EE}-V_{EB})}{R_E}=0,93 \text{mA} $$

$$ I_E=I_B+I_C=\frac{I_C}{\beta}+I_C$$

Therefore,
$$ I_C=I_E \times \frac{\beta}{(1+\beta)}=0.921 \text{mA} $$