 Solution

Electromagnetism gauges the vector flavour transformation $U\mapsto e^{ie\alpha Q}U{e}^{-ie\alpha Q}$, so covariance requires $D_{\mu }U={\partial }_{\mu }U-ie{A}_{\mu }[Q,U]$. The identity part of $Q=I/6+{\sigma }_3/2$ drops out of the commutator. Writinggives $[Q,\pi {]}_{12}={\pi }_{12}$ and $[Q,\pi {]}_{21}=-{\pi }_{21}$. Hence ${\pi }^{+}$ has electric charge $+e$, ${\pi }^{-}$ has charge $-e$, and ${\pi }^0$ is neutral.