 Solution

The transformation of the holomorphic strong-coupling scale is fixed by the mixed $SU(2{)}^{2}U(1)$ anomaly. The baryon-number contributions of $\Phi$ and $\tilde{\Phi }$ cancel, so ${\Lambda }^{b_0}$ is neutral under $U(1{)}_B$. Under $U(1{)}_A$, all $2N_f$ doublets have charge one and Dynkin index one, giving axial charge $2N_f$. For the R-symmetry, each matter fermion has charge $(1-2/N_f)-1=-2/N_f$, so the matter contribution is $-4$; the gaugino has R-charge one and contributes $I(\mathrm{adj})=4$, giving zero total anomaly. Thus