Biasing circuit with mosfet: am I missing information or misreading it?
The current through both the MOSFETs are same. Also it is given that both the MOSFETs are in saturation region. Therefore we can write:
$$I_{D1} = I_{D2}= I_{D} = k(V_{GS} - V_t)^2 = 20 \mu A $$
on solving we get \$V_{GS}\$ should be \$2.4V\$  for both the MOSFETs, if \$ 20 \mu A\$ has to flow through both of them.



Now for M2, 
$$V_{GS} = 2.4V$$
$$ \implies V_B - V_A = 2.4 V $$
Given \$V_B\$ is 5V. 

$$ \implies 5 - V_A = 2.4 V $$
$$\implies V_A = 2.6V$$

For M2 to enter saturation, \$ V_{DS}\$ of M2 should be atleast \$ V_{GS} - V_t\$
$$ V_{DS_{min}}= 2.4 - 1 = 1.4 V $$
At this point, \$ V_{DS} \$ of M1 will be:
$$V_{DS(M1)} = V_A = 2.6V &gt; 1.4V$$

So M1 will also be in saturation, well saturated over the minimum value of \$1.4V\$.

Now from the circuit we can express \$V_{DS}\$ of M2 as
$$V_{DS} = V_o - V_A $$
$$\implies V_o = V_A + 1.4 = 4 V$$

If you look closely, the node voltage \$ V_A \$ decides the minimum output voltage. When M2 entered the saturation region (Boundary condition), M1 had already been saturated way before. So your assumption of having boundary conditions in both the MOSFETs at the same time, went wrong. And hence the calculations too.