Low Pass Filter – How to Calculate the Cutoff Frequency

You have two cascaded filters, but due to the resistive dividers they are not the same. If they were, it would have been as Bimpelrekkie and jonk said in the comments. Either way, the frequency would be a geometric mean of the two stages. Consider two, generic, all-pole, 2nd order transfer functions:

$$\begin{align} H_1(s)&=\dfrac{\omega_1^2}{s^2+\dfrac{\omega_1}{Q_1}s+\omega_1^2}\tag{1} \\ H_2(s)&=\dfrac{\omega_2^2}{s^2+\dfrac{\omega_2}{Q_2}s+\omega_2^2}\tag{2} \\ H_1(s)\cdot H_2(s)&=\dfrac{\omega_1^2\omega_2^2}{s^4+\Biggl(\dfrac{\omega_1}{Q_1}+\dfrac{\omega_2}{Q_2}\Biggr)s^3+\Biggl(\omega_1^2+\dfrac{\omega_1\omega_2}{Q_1Q_2}+\omega_2^2\Biggr)s^2+\omega_1\omega_2\Biggl(\dfrac{\omega_2}{Q_1}+\dfrac{\omega_1}{Q_2}\Biggr)s+\omega_1^2\omega_2^2}\tag{3} \\ \end{align}$$

To find out \$\omega\$ for a 2nd order, you would use \$(\omega^2)^{\frac12}=\sqrt{\omega^2}=|\omega|\$. Simmilarly, for an Nth order, there would be a \$\omega^N\$ term which will be calculated as \$(\omega^N)^{\frac1N}\$. In this case, it would be:

$$(\omega_1^2\omega_2^2)^{\frac14}=\omega_1^{\frac24}\omega_2^{\frac24}=\sqrt{\omega_1\omega_2}\tag{4}$$

Which is the geometric mean. Note that this would be the corner frequency as defined by the transfer function, or the frequency when the phase reaches half its final (asymptotic) value.

For your case, that particular topology is that of a Sallen-Key and the transfer function for one stage is (considering the notations of the first stage):

$$\begin{align} H(s)&=\dfrac{\dfrac{K}{R_1R_2C_1C_2}}{s^2+\Biggl(\dfrac{1}{R_1C_1}+\dfrac{1}{R_2C_1}+\dfrac{1-K}{R_2C_2}\Biggr)s+\dfrac{1}{R_1R_2C_1C_2}}\tag{5} \\ K&=1+\dfrac{R_3}{R_4}\tag{6} \end{align}$$

The two corner frequencies are the same, ~189.8 Hz, but their quality factors differ. In the picture that you're showing, the 2nd stage is not a stable filter (the phase of V(o2), blue trace, goes positive), so for the overall response I modified R7 to be 6k, only (to get within a stable region):

As you can see, the corner frequency for V(o) is around the same place as the other two, except that the quality factor of the 2nd stage is higher and influences the overall response.