Trouble with voltage BJT circuit in active mode
I agree with your work regarding estimating the current in \$R_4\$ as \$I_{R_4}\approx 19.3\:\textrm{mA}\$. So your analysis there is fine. Yes, \$Q_3\$ and \$Q_4\$ make up a current mirror, so you'd expect a similar sinking current in the collector of \$Q_3\$. (I'm also assuming here that we do not need to worry about the Early Effect, which in this case would otherwise probably be an issue.) \$Q_3\$'s collector should have a compliance voltage down to perhaps \$-19.3\:\textrm{V}\$. (That won't really matter because the emitter of \$Q_2\$ will keep it from getting anywhere below \$-1\:\textrm{V}\$.)

Now, you need to assume that \$Q_2\$ is active. This means \$V_{B_2}=V_{C_2}=0\:\textrm{V}\$. (Easily seen in the circuit.) Until this point, \$Q_2\$ will be saturated and a lot of the current mirror's sinking current will also come from the base of \$Q_2\$. But right at the point where \$Q_2\$ becomes active we can assume \$\beta=100\$ and its base current will be "normal." This also means that the current in \$R_3\$ must be \$I_{R_3}=\frac{10\:\textrm{V}-0\:\textrm{V}}{R_3=1\:\textrm{k}\Omega}=10\:\textrm{mA}\$. So the emitter current will be the collector current plus 1% for the base current, or \$I_{E_2}=10.1\:\textrm{mA}\$.

\$Q_1\$'s collector has the remaining difference of \$I_{C_1}=19.3\:\textrm{mA}-10.1\:\textrm{mA}=9.2\:\textrm{mA}\$. With \$\beta=100\$, the base current for \$Q_1\$ must be \$I_{B_1}=\frac{9.2\:\textrm{mA}}{100}=92\:\mu\textrm{A}\$. That has to be supplied through \$R_1\$. So the voltage across it must be \$\mid\:V_{R_1}\mid\:=92\:\mu\textrm{A}\cdot 100\:\textrm{k}\Omega=9.2\:\textrm{V}\$.

The only remaining question is, "what is the voltage at the base of \$Q_1\$?" Well, we don't actually know exactly. But it must be very close to \$0\:\textrm{V}\$. But I think your problem assumes all the \$V_{BE}=700\:\textrm{mV}\$; and in any case we know that \$Q_1\$ itself would just be barely active if its base were at \$0\:\textrm{V}\$. So I think it's reasonable here to argue that the problem wants you to take \$V_{B_1}=0\:\textrm{V}\$. And there isn't anything much here to argue strongly against that assumption. So there it is.



So this means the answer is, \$Q_2\$ becomes active when \$V_i\ge 9.2\:\textrm{V}\$.