Answer by periblepsis for What should be the collector current for Q1 in this Razavi problem if I assume the emitter and collector currents to be equal?

Am I correct to assume Ie equals Ic as the beta is quite large here?

Not unless you've been told to ignore it. And since \$\beta\$ has been specified as a finite value, that would be a clear indication to me that you are not supposed to ignore the base current. I could be over-ridden by other instructions you may have been given. But that's how I'd interpret what you've shown.

Is there any mistake in how I have approached ?

Perhaps. I started out trying and then decided that I wasn't going to go any further trying to read your hand-writing. Since I had to give up on that, I just decided to use my own approach and let you decide where your errors are at, if any.

1. \$Q_1\$ has twice the saturation current as \$Q_2\$ and their bases and emitters are tied, so they both have the same \$V_{\small{BE}}\$. If I assume that \$Q_1\$ isn't saturated by the presence of \$R_{\small{C}}\$, then given also the fact that there's no Early Effect I can replace \$Q_1\$ and \$Q_2\$ by a single transistor with a saturation current of \$7.5\times 10^{^{-16}}\:\text{A}\$.
2. \$V_{\small{TH}}=1.2\:\text{V}\$ and \$R_{\small{TH}}=6.24\:\text{k}\Omega\$.
3. \$I_{\small{B}}=\frac{V_{\small{TH}}-V_{\small{BE}}}{R_{\small{TH}}+\left(\beta+1\right)R_{\small{E}}}\$. Unfortunately, I don't know (yet) what \$V_{\small{BE}}\$ is. I could solve it using the product-log function (branch-0 of LambertW.) Otherwise it must be solved iteratively. Start with a guess of \$V_{\small{BE}}\approx 700\:\text{mV}\$ to find \$I_{\small{B}}\approx 10.72\:\mu\text{A}\$.
4. At standard temp of \$27^\circ\text{C}\$\$V_T=25.865\:\text{V}\$. So adjust and find \$V_{\small{BE}}=25.865\:\text{mV}\cdot\ln\left(1+\frac{\beta\,\cdot\, 10.72\:\mu\text{A}}{7.5\times 10^{^{-16}}\:\text{A}}\right)\approx 724\:\text{mV}\$.
5. Return to step #3 above and find \$I_{\small{B}}\approx 10.21\:\mu\text{A}\$ and to step #4 above and find \$V_{\small{BE}}\approx 722.7\:\text{mV}\$.
6. Return to step #3 above and find \$I_{\small{B}}\approx 10.23\:\mu\text{A}\$ and to step #4 above and find an unchanged \$V_{\small{BE}}\approx 722.7\:\text{mV}\$. That's where things stop. (LambertW would have solved things at step #3.)

At this point I know enough.

\$I_{\small{C}_1}=\frac23 \beta \cdot 10.23\:\mu\text{A}\approx 682\:\mu\text{A}\$ and \$I_{\small{C}_2}=\frac13 \beta \cdot 10.23\:\mu\text{A}\approx 341\:\mu\text{A}\$.

I can also confirm that \$Q_1\$ is not saturated by the presence of \$R_{\small{C}}\$. So the approach holds.

Simulation shows:

Since I didn't retain more than a few digits of precision in my hand-work, I'd call that a match.

Another approach would be to just recognize that \$I_{\small{B}_1}=2\cdot I_{\small{B}_2}\$. So that the KVL equation becomes \$V_{\small{TH}}-3\cdot I_{\small{B}_2}\cdot R_{\small{TH}}-V_{\small{BE}}-3\left(\beta+1\right)\cdot I_{\small{B}_2}\cdot R_{\small{E}}=0\:\text{V}\$. Then just work things out from the point of view of \$Q_2\$, knowing that the collector current for \$Q_1\$ will be twice that much.