Required Emitter Resistance \$R_E\$ for the Maximum Emitter Current for a BJT
7 years late, but this was a fun one to do some forensic math on! And it actually may be a bit of a trick question! The trick comes down to that you're given an input current \$I_B\$, not an input voltage.
First: The error was on line 3 when you calculated \$I_E\$. You actually found \$I_C\$ instead. You have to add \$I_B\$ to that to get \$I_C\$.
Here's the shortcut solution. If the BJT is in the active mode:
$$
I_E=(\beta+1)I_B
$$
That means that it doesn't matter what \$R_E\$ is, the current will just be \$101\times I_B\$. We can explore this a little bit more thinking about Ebers Moll



simulate this circuit &ndash; Schematic created using CircuitLab
Let's say we start off with \$I_B=0\$. At this point CE is positive, reverse biasing D2 - so no current flows anywhere. This is in cutoff. It doesn't matter what \$R_E\$ is in this case either.
Let's turn on \$I_B\$ a little bit. D1 is forward biased and D2 is still reverse biased (so it's in the active mode). The current through D1 is \$I_{D1} = I_B+\alpha_F I_{D1}\$. Solving for the current through D1 is \$I_{D1}=I_E= I_B/ (1-\alpha_F)I_B = (\beta +1)I_B\$. Again, the value of \$R_E\$ does not matter.
Ok, so our BJT is in the active state. That means:
$$
V_BE=0.7\\
I_C=\beta I_B\\
V_o = V_+ - \beta I_B R_C=V_C\\
V_E = (\beta+1)I_B R_E-V_+\\
V_B \approx 0.7 - V_+ +(\beta+1)I_B R_E\\
V_{BE} \approx 0.7\\
V_{BC} \approx -2V_+ + 0.7 + (\beta+1)I_B R_E + \beta I_B R_C
$$
That means that at some point \$I_B\$ gets large enough that the \$V_{BC}\$ becomes positive, and our diode D1 becomes forward biased. We are now in saturation mode. At that point, no matter how much we increase IB, \$I_C\approx I_E\$. \$\beta\$ is effectively reduced, but it's still basically \$I_E=(\beta_{reduced}+1)I_B\$.
So it's not exactly a trick question, but it's a question about controlling a BJT with current instead of voltage.