I'm not quite sure what the question means. I need to explain what would happen if I removed the \$1.0 \mu F\$ blocking capacitor from this circuit. Howver, I'm not sure what the "DC bias voltage of the emitter" is. Is it simply the voltage as measured from base to emitter? This is for a 2N3904 transistor and an emitter-follower setup.
They probably mean the voltage on the emitter relative to ground. It should not matter though.
This is a homework questions so I won't give you the exact answer, but think about what happens after a long time has passed to the current through a capacitor and how that would affect the voltage.
If Trevor/KeleMP are correct that they mean to replace it by a short (we can't see the actual wording in the question), the analysis becomes straightforward and you can discover why you would want a blocking capacitor in there. On looking up the actual wording, I think they are correct: