Finding the angular cutoff frequency
The cut-off frequency is where the gain is 3dB below the passband gain, \$=K\$, say. Assuming your equation relates to voltage gain, -3dB equates to \$\dfrac{1}{\sqrt2}K\$

The equation represents a 2nd order low-pass filter and the passband gain is found when \$\omega=0\$. This gives a passband gain of \$K=10\$, hence the cut-off frequency can be determined by setting \$|H(j\omega)|=\dfrac{10}{\sqrt2}\$