How to find 3 dB bandwidth of any transfer function?
the "3 dB bandwidth" is more precisely called the "-3.01 dB bandwidth" and even more precisely called the "half-power bandedge".

so begin with your transfer function 

$$ H(s) = \frac{b_0 + b_1s + b_2s^2 + ... + b_N s^N}{a_0 + a_1s + a_2s^2 + ... + a_N s^N} \triangleq \frac{P(s)}{Q(s)} $$

evaluate this at an arbitrary frequency \$\omega\$ on the imaginary axis on the \$s\$-plane.

$$ H(j\omega) = \frac{P(j\omega)}{Q(j\omega)} $$

that represents voltage gain (and phase shift) frequency response.  for power gain, it's the magnitude-squared of the frequency response:

$$ |H(j\omega)|^2 = \left| \frac{P(j\omega)}{Q(j\omega)} \right|^2 $$

evaluate the magnitude squared at the "center" of your passband.  let's say it's a low-pass filter (LPF) as is the example in the question above.  the center of the passband (when you also include negative \$\omega\$) is at DC or \$\omega=0\$

$$ |H(j0)|^2 = \left| \frac{P(0)}{Q(0)} \right|^2 $$

the -3 dB frequency is at the frequency which results in half of the power as in the center of the passband (in this case, at DC).

$$ |H(j\omega)|^2 = \frac12 |H(0)|^2 $$

or

$$ \left| \frac{P(j\omega)}{Q(j\omega)} \right|^2 = \frac12 \ \left| \frac{P(0)}{Q(0)} \right|^2 $$

or

$$ 2 \ |Q(0) \ P(j\omega)|^2 = |P(0) \ Q(j\omega)|^2 $$

solve for \$\omega\$ and you have your -3 dB bandwidth.