= Solution
Take <metric signature> $(+,-,-,-)$ and the supplied covariant <electromagnetic field tensor>, so $F_{0i}=E_i$ and $F_{ij}=-\epsilon_{ijk}B_k$. Raising both indices reverses the time-space components but leaves the spatial components unchanged. Its complete contraction is therefore
$$
F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2).
$$
Define the <dual electromagnetic field tensor> by $\widetilde F^{\mu\nu}=\tfrac12\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$, with $\varepsilon^{0123}=+1$. Direct contraction gives
$$
\boxed{E^2-B^2=-\frac12F_{\mu\nu}F^{\mu\nu},\qquad
\mathbf E\cdot\mathbf B=-\frac14F_{\mu\nu}\widetilde F^{\mu\nu}.}
$$
These <electromagnetic field invariants> are complete tensor contractions. The second is a <pseudoscalar>: it is invariant under proper <Lorentz transformations> and changes sign under spatial inversion.
For $T^{0i}$ the metric trace term vanishes. Since $F^0{}_j=E_j$,
$$
F^0{}_{\rho}F^{i\rho}=\sum_jE_j(-\epsilon_{ijk}B_k)=-(\mathbf E\times\mathbf B)_i.
$$
Consequently the <Poynting vector> is exactly the time-space row in the stated $c=1$ units:
$$
\boxed{(T^{01}_{\rm em},T^{02}_{\rm em},T^{03}_{\rm em})
=\frac1{\mu_0}\mathbf E\times\mathbf B.}
$$
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