maximum possible common mode voltage for npn
You need Vout = 1V when you have Vin = 10mV. These values are fully-differential. That means at single-ended input of 5mV you have to got a 500mV change in output.

First get the g_m of one Q1 by:

$$g_m = \cfrac{I_c}{2\cdot V_{Th}} = \cfrac{1mA}{2\cdot 25\,mV} = 20 \, mS\, ,$$

assuming that the Q1 runs at room temperature. Therefore, when we have the maximum input voltage of 5mV, you get:

$$\Delta  I_C = g_m\cdot \Delta v_{BE} = 20\, mS \cdot 5\, mV = 0.1\, mA\, .$$

To get a value of R_c, you have to remind that the voltage drop has to be 500mV:

$$R_C = \cfrac{500\, mV}{\Delta I_C}\, = 5\,k\Omega .$$

The branch with the higher current in the differential amplifier sets the maximum allowable common mode input voltage.

Looking at the circuit diagram below, you can set a voltage loop of the input.









simulate this circuit &ndash; Schematic created using CircuitLab



$$V_{CC} - R_C \cdot \Bigg(\cfrac{I}{2} + \Delta I_C \Bigg) - V_{BC}\ge V_{in\,\text{Common mode}} + V_{in\, \text{diff mode}} \\
5\, V  -  5\, k \cdot 0.6 \, mA + 0.4\, V\ge V_{in\,\text{Common mode}} + 5\, mV \, \\
V_{in\,\text{Common mode}} \le 2.395 \,V\quad .$$

NOTE:When the voltage at the collector is higher than the base voltage, the transistor Q1 keeps in the triode  (linear, active) region. In this textbook and as simulation shows, the base-collector diode can open until a voltage of 400mV is reached, without influencing the gain stage.

Your textbook has forgotten to include the 5mV of the max. differential input voltage. In case the gain is high, it doesn't change much, however, when the gain is low (e.g. 2), it matters! If the gain is 2, and you want to have a max. single-ended output swing of 500mV, you get a V_differential/2 of 250mV! , wich you have to add to your max. allowablecm input range.
Most textbook ignore that case!

Here is the simulation result

The black curve is the collector voltage of Q1, which starts to increase at a common mode input voltage of 2.45V