Could you get it?

Yes, could you explain some of the terms and more importantly you state that $$ E = \int T_{00}d^{3}\mathbf r = \int \frac{1}{8 \pi}\left( \hat{\mathbf B}^{2} + \hat{\mathbf E}^{2}\right)d^{3}\mathbf r. $$ but what does $E$ denote? I'm looking for $E_{p}=f(E_{w})$

@J.D'Alembert : $E$ here means the full energy of EM field which is given from lagrangian formalism. Here $T_{00}$ is equivalent to hamiltonian density. $\hat{\mathbf E}, \hat{\mathbf B} $ are respectively the electric field strength and the magnetic field induction.

@AndrewMcAddams Ok, so the question is now, how does one relate the energy of sets of photons to $E$?

@J.D'Alembert : I'll write some calculations and thinking in a few time into the answer.

@New_new_newbie : classical fields as quantum ones are localized on infinity.

@AndrewMcAddams May I ask what $\mathbf r$ is?

@J.D'Alembert : $\int d^{3}\mathbf r = \int dV$ here denotes integration over field lozalization. Since we discuss free case (EM field in vacuum) field localization is infinity.

@J.D'Alembert : I afraid that in this cumbersome and non-interested math is possible to sink, but I've finished to derive the relation.

@AndrewMcAddams Sorry that it took me this long to accept it my internet was disconnected and thank you for all the work that you put into your answer.

@AndrewMcAddams Could you explain the notation used in $$\sum_{\lambda}\int d^{3}\mathbf p E_{\mathbf p}\left( \hat{a}_{\lambda}^{\dagger}(\mathbf p)\hat{a}_{\lambda}(\mathbf p) + \delta (0)\right)$$?

@J.D'Alembert : here $\lambda$ means polarization (photon has 2 independent polarizations), $E_{\mathbf p} = \omega_{\mathbf p}$ denotes energy of photon with momentum $\mathbf p$, $\delta (0)$ arises because of I have made intercharge $$ \hat{a}_{\lambda}(\mathbf p)\hat{a}_{\lambda}^{\dagger}(\mathbf p) = \delta (\mathbf p - \mathbf p) + \hat{a}^{\dagger}_{\lambda}(\mathbf p)\hat{a}_{\lambda}(\mathbf p). $$ Usually we neglect by this summand because it is interpreted as vacuum energy and we can shift the energy level by this constant.

@AndrewMcAddams if $\delta (0)$ is the dirac delta function wouldn't it mean the solution is infinite as it approaches infinity?

@J.D'Alembert : In a few words an explanation is following. QFT is (in some sense) ill, because it has singularities, which are hidden in the commutators. At the same time it means that we can modify its primary objects (lagrangian, for example) by some ways with only one requirement: these changes must correspond to physics, because this infinity is the consequence of only mathematical description. We see that infinity in expression $(6)$ is caused only by commutators. So we may throw this constant away.

@J.D'Alembert : it seems yes, because energy is positive definite quantity.

@AndrewMcAddams Last questions, is it possible to rearrange equation (6) for $E_{\mathbf p}$ and if so how?

@J.D'Alembert : what did you mean?

Could you solve $$\hat{H}=\sum_{\lambda}\int d^{3}\mathbf p E_{\mathbf p}\left( \hat{a}_{\lambda}^{\dagger}(\mathbf p)\hat{a}_{\lambda}(\mathbf p) + \delta (0)\right)$$ for $E_{\mathbf p}$ (ignoring the constant)?

@J.D'Alembert : in some sense you can solve this equation for $E_{\mathbf p}$ by acting of $\hat{H}$ on some one-particle state $| \mathbf p , \lambda\rangle$: then $$ \hat{H}| \mathbf p , \lambda\rangle = E_{p} | \mathbf p , \lambda \rangle . $$