Circuit Analysis: Obtaining Close Loop OP - AMP Transfer function
I believe your mistake is to assume 
$$\ B = G(s) = \frac{R2}{R1}+1 = \frac{1000k\Omega}{1000k\Omega}+1 = 2$$
B is actually 1/2, as it is the output voltage divided by 2, that is 
$$\ B = G(s) = \frac{R1}{R1+R2} $$
Where the ratio comes from the \$R1,R2\$ voltage divider.

With this value of B, you would obtain
$$\ CL(0) = \frac{31.62}{16.81} = 1.85$$