Here's what I'm doing, but I'm not sure if this is correct.
Furthermore, I think $c^2$ is the speed of light, however is $c^2 = (\mu_0\epsilon_0)^{-1}$ as follow?
$$\nabla \times (\nabla \times \vec{E}) = \nabla \times(- \frac{\partial\vec{B}}{\partial t})$$
$$\nabla(\nabla \cdot\vec{E})- \nabla^2\vec{E} = - \frac{\partial}{\partial t}(\nabla \times \vec{B})$$
$$-\nabla^2\vec{E} = -\frac{\partial}{\partial t}(\mu_0\epsilon_0 \frac{\partial\vec{E}}{\partial t} + \mu_0\vec{J})$$
$$\nabla^2\vec{E} = \frac{\partial^2\vec{E}}{\partial t^2}\mu_0\epsilon_0$$
$$\Delta \vec{E} = \frac{1}{c^2} \frac{\partial ^2\vec{E}}{\partial t^2}$$
