3 dB bandwidth of a filter
From the given transfer function you can see that the filter has a low pass character. The gain at DC is given by $H(1)=1$. This is also the maximum value of $|H(e^{j\omega})|$. So you can try to solve
$$\big|H(e^{j\omega_c})\big|^2=\frac{\max_{\omega}|H(e^{j\omega})\big|}{2}=\frac12\tag{1}$$
because this equation defines the frequency $\omega_c$ at which the filter gain equals $1/\sqrt{2}$, which is the definition of the $3$ dB cut-off frequency.
Your calculation of the magnitude of $H(e^{j\omega})$ is correct, so you can just continue by solving $(1)$.
As a final note, there are values of $a$ for which the $3$ dB cut-off frequency doesn't exist because the filter never reaches an attenuation of $3$ dB. You can try to find that limit of $a$ as a bonus exercise.