Your amplifier is not a 2 stage one strictly speaking. It has only one high-impedance node (the output) and the diff-pair itself does not provide a lot of gain since it is loaded by diode connected transistors. Compensation is also done differently compared to the two-stage Miller compensated OTAs. Here you compensate with the load capacitor. This being said, 100mS of transconductance seems quite a lot for 0.35u CMOS. But, here's how I would think about it. I also assume that by transconductance you mean the output current of the OTA vs the input diff voltage. Whatever current you get from the output splits in 2 and gets divided by the ratio of M2-M3 (M4-M5). So, the current the diff pair has to produce is smaller by that ratio - say 10 times. The transonductance of the input transistor is then 10mS. Since you don't say anything about BW requirements, I assume it is not important. Then, you can fix the Vov of the input devices which is something like 2/gm/Id and since you know how much gm you want, you find Id. Choose L and find W. Or you can fix the current you want to consume, then find gm/Id and for a given L you can get Id/W and from here W. In all this I assume you are familiar with the gm/Id design methodology. If not then for the above two approaches you use the square low transistor formulas. I also want to mention that many times for this topology a mirror ratio of 10 (M2-M3; M4-M5) is a lot for many reasons. 2-5 is more likely. Then you diff pair transistors have to provide bigger transconductance.