Voltage for transistors in saturation
Your circuit doesn't really make any sense, in my opinion. Chances are, it was created almost at random and without any "intelligent actor" behind it. You'll see why I have that opinion in a moment.

Let's redraw that crazily laid out schematic, to start.







simulate this circuit &ndash; Schematic created using CircuitLab

Let's analyze the schematic first by assuming that \$Q_2\$ is completely off (\$V_I=20\:\textrm{V}\$) and see where that leaves things.

Here's the KCL nodal equation for \$V_X\$:

$$\begin{align*}
\frac{V_X}{R_3} + \frac{V_X}{R_6} &amp;= I_E + \frac{700\:\textrm{mV}}{R_3}\tag{Eq. 1}
\end{align*}$$

Where we know this from KVL:

$$\begin{align*}
V_X&amp;=20\:\textrm{V} -I_E\cdot R_2-.7-\frac{I_E}{\beta_1+1}\cdot R_5\label{vx}\tag{Eq. 2}
\end{align*}$$

Let's see what assuming \$\beta_1=100\$ gets us:

$$\begin{align*}
I_E &amp;\approx 1.916\:\textrm{mA}\\\\
V_X &amp;\approx 8.772\:\textrm{V}
\end{align*}$$

From here, we can work out that \$I_{R_3}=\frac{V_X-700\:\textrm{mV}}{R_3}= 161.44\:\mu\textrm{A}\$. Given your own computation that \$I_{C_3}=3.96\:\textrm{mA}\$, this already works out to \$\beta_3\approx 24.5\$ and that is already well below your nominal value of \$100\$.

So it's clear, now, that without any contribution from \$Q_2\$ the circuit already puts \$Q_3\$ in at least a shallow saturation situation.

Okay. So now what? Well, \$Q_2\$ won't contribute anything until its base voltage is at least \$700\:\textrm{mV}\$ below the above-computed value for \$V_X\$. So \$V_I\le 8\:\textrm{V}\$, roughly speaking. As you imagine the value of \$V_I\$ declining towards (and perhaps below ground), \$Q_2\$ will start pulling away current and thereby gradually moving \$Q_3\$ out of saturation.

That will happen about when \$\beta_3=100\$, or when \$V_X\approx 2.7\:\textrm{V}\$. From \$\ref{vx}\$ above, we can easily work out that this happens when the emitter current is \$I_{E_1}\approx 3\:\textrm{mA}\$. Subtracting \$I_{B_3}\approx 40\:\mu\textrm{A}\$ and \$I_{R_3}\approx 540\:\mu\textrm{A}\$, this leaves about \$I_{C_2}\approx 2.42\:\textrm{mA}\$. As \$V_{CE_2} \approx 2.7\:\textrm{V}\$, \$Q_2\$ is not in saturation at this time so the full \$\beta_2=100\$ can be used and we figure \$I_{B_2}\approx 24.2\:\mu\textrm{A}\$. We know that \$V_{B_2}\approx 2\:\textrm{V}\$, so this means that \$V_I\le 2\:\textrm{V} - 100\:\textrm{k}\Omega\cdot 24.2\:\mu\textrm{A}\approx -420\:\textrm{mV}\$.

So now we know approximately where \$Q_3\$ leaves saturation and enters into active mode: \$V_I\le -420\:\textrm{mV}\$.



\$Q_3\$ is in saturation for all positive values of \$V_I\$ and computing the point where it leaves saturation is a bit nuanced. This is why I think the whole question was made up without much attention given to it. (It might be a good question if you are somewhat advanced in DC analysis.)