 Solution

For $H_0=E_2{B}_2^{†}{B}_2$, the eigenspaces of energies $0$ and $E_2$ are each two-dimensional because $n_1$ is free. Degenerate perturbation theory requires diagonalizing $E_1{B}_1^{†}{B}_1$ within each space. It is already diagonal in the product basis, with eigenvalues $0,E_1$. Therefore the first-order eigenvectors remainand their energies are $0,E_1,E_2,E_2+E_1$, respectively. These are the exact answers, so all higher corrections vanish.