Output resistance of MOSFET circuit
Probably the correct way to draw your circuit is by connecting the tops of the 10 k drain resistors together and to ground, as I assume they are in fact connected to the VDD rail, which however has no place in a small signal model and should be considered as ground.
The output should be totally disconnected and rather be connected to the output terminal where the load goes, as now it is short circuited to ground (for AC).
Also, g is expressed in \$ (m)A/V \$, not \$ (m)A/V^2 \$.

Big Edit:

Now I want to know exactly what is the case, and will try to calculate open output voltage and short-circuited output current in the presence of input signal \$V_i\$ and divide the two in order to get the output impedance.

So in LTSpice I drew the original circuit in a small-signal representation which is void of all things DC, and this is what it looks like: 

 
In which \$R_8\$ should read \$R_2\$ and the '\$R\$' in the same is \$10 k \, \Omega\$.
If the mosfets in this circuit are replaced by a simple equivalent circuit with only \$g\$ and \$R_o\$, then we get the circuit below: 

and in this representation we can replace the current sources and internal resistances with voltage sources as in the picture below:


Going from left to right through the circuit we can clearly see that the current through \$R_3\$ is $$I_{R_3} = g R_o U_{gs_1} / (R_1 + R_3 + R_5) $$ 

Further, $$U_{gs_1} = U_{g_1} - R_3 I_{R_3} $$ so that $$ I_{R_3} = g R_o (V_i - R_3 I_{R_3})/(R_1 + R_3 + R_5) $$. Rearranging this gives $$ g R_o V_i / (R_1 + R_3 + R_5) = I_{R_3} \{R_1 + (1 + g R_o)R_3 + R_5\}/(R_1 + R_3 + R_5) $$ from which we find 
$$ I_{R_3} = g R_o V_i / \{R_1 + (1 + g R_o)R_3 + R_5\} $$

Now, for the second mosfet. It's gate voltage we find from $$ U_{g_2}=-R_1I_{R_3} = -gR_oR_1V_i/\{R_1+(1+gR_o)R_3+R_5\}$$ and we find the current through \$R_6\$ 
$$ I_{R_6}=-(gR_o)^2R_1V_i/\{R_1+(1+gR_o)R_3+R_5\} $$

Now up to the third mosfet we can write its gate voltage
$$ U_{g_3} = -R_2I_{R_6} = (gR_o)^2R_1R_2V_i/\{R_1+(1+gR_o)R_3+R_5\} $$
Now we also have
$$ U_{gs_3} = U_{g_3} - R_4I_{R_4} $$
and finally we can now calculate both open output voltage and short circuit output current. Namely if the output is open, \$I_{R_4}=0\$ and \$U_{gs_3}=U_{g_3}\$ and we can write the open output voltage as
$$ U_{out_{open}} = -(gR_o)^3R_1R_2V_i/\{R_1+(1+gR_o)R_3+R_5\} $$
and if the output is short circuited then \$U_{out}=0\$ and we can write
$$ I_{out} = -I_{R_4} = -g(U_{g_3}-R_4I_{R_4}) = -gU_{g_3}/(1+gR_4) $$
which, rearranged, gives us the short circuit output current
$$ I_{out_{short}} = -(gR_o)^3R_1R_2V_i/\left[\{R_1+(1+gR_o)R_3+R_5\}R_o(1+gR_4)\right] $$
the form of which looks eerily similar to that of \$U_{out_{open}}\$ and the quotient of the two becomes simply 
$$ Z_{out} = \frac{U_{out_{open}}}{I_{out_{short}}} = (1+gR_4)R_o $$
With \$g = 4mA/V\$ and \$R_o=100\, k\Omega\$, this becomes
$$ Z_{out} = (1+4.10^{-3}.90).10^5 = 136 \, k\Omega $$