EDIT: My answer was updated. In the first version, by mistake I have assumed that the common-mode input resistance - instead of common-mode gain - was asked.
Rather than providing the complete formula for the common-mode gain, I like to describe the way how to derive the desired result. This should enable you to solve the task by yourself.
As you know, gain is the output-to-input ratio. For common mode signals it would be, in principle, sufficient to consider one input only because both transistors are doing the same. Therefore, looking into one input only you have to analyze a simple common-emitter stage with negative feedback caused by the resistance in the common emitter leg. This feedback effect, of course, drastically decreases the common mode gain.
In the present case, feedback is provided not by an ohmic resistor (Re) but by a differential (dynamic) resistance (re). This feedback resistance re has a finite value (output resistance of a bjt identical to the inverse slope of the Ic=f(Vce) characteristic of the transistor in the common emitter leg).
Hence, the known gain formula for a common-emitter stage with resistive feedback can be applied. However, there is one important point which must be not overlooked:
The current change to be observed in the feedback resistor for common-mode operation is caused by BOTH (equal) input signals. Hence, the corresponding voltage change causing the feedback voltage is TWICE the value belonging to one transistor only. This effect is easily covered if we replace "re" in the formula for the input resistance by the expression "2re". That`s all.