The givens in your problem statement are contradictory.
If Vbe is 0.68 V, then, as you say, the current through R1 is 0.1 mA.
Given the current source at 1.1 mA, then the emitter current must be 1.0 mA.
But you were also told that Ic is 1 mA.
There's a variant of KCL that says that if we draw a closed curve across our circuit, then the sum of all the currents into the enclosed region must be 0.
So if we draw a circle that encloses the BJT and no other elements of the circuit, the total current into that circle must be 0. This means
Ic + Ib + Ie = 0
if all of the currents are taken with a positive sign indicating current going in to the device.
Since Ic is 1 mA, and Ie is -1 mA, then we must have Ib = 0 mA.
Which violates the characteristic equations for the device (Ic = β Ib).
At least one of the "givens" in the problem statement must be rejected. Either Ic isn't really 1 mA, or R1 isn't really exactly 6.8 kOhm, or IS isn't really exactly 1.1 mA or ...
I suggest discarding the assumption that Ic is 1.0 mA and solve the problem from there.