From Quantum Field Theory by Franz Mandl and Graham Shaw page 4.
When we are expanding the vector potential as a Fourier series;
$\renewcommand{\vec}[1]{\mathbf{#1}}\vec{A}(\vec{x},t) = \sum\limits_{k}\sum\limits_{r}(\frac{\hbar c^2}{2V\omega_k})^{1/2}\vec{\varepsilon_r}(\vec{k})[a_r(\vec{k},t)e^{i\vec{k}\cdot\vec{x}}+a_r^*(\vec{k},t)e^{-i\vec{k}\cdot\vec{x}}]$
I did not understand how we can determine the constant term $(\frac{\hbar c^2}{2V\omega_k})^{1/2}$. Thanks in advance.
That factor is there because they quantize the field in a box and not in the continuum. When you take the continuum limit one gets $$ \int d^3k \rightarrow \sum_k \sqrt{\frac{\hbar c^2}{V}} $$ by dimensional analysis.
The $1/\sqrt{2\omega_k}$ is related to Lorentz invariance. A factor of $1/2\omega_k$ makes the measure $ d^3k/2\omega_k$ Lorentz invariant (see this post) and the remaining $\sqrt{2\omega_k}$ goes with the creation operators so that the definite momentum states $$ |k\rangle = \sqrt{2\omega_k} a^\dagger(k)|0\rangle $$ have a Lorentz invariant measure too, as @Hunter explains.
