 Solution

In left-handed variables, both ${\psi }_L$ and ${\tilde{\psi }}_R$ have $U(1{)}_X$ charge $-1$. Cancellation of the mixed $SU(2{)}_{\mathrm{gauge}}^{2}U(1{)}_X$ anomaly requireswhere $I(2)=1$ and $I(\mathrm{adj})=4$. Therefore