Let M be an anti-self dual and Einstein 4-manifold with scalar curvature s. Then the Ricci tensor $c_Z$ of the twistor space is given by $$c_Z (E,E) = (s/4 - t(s/12)^2) \|X\|^2 + (1+ (ts/12)^2)\|V\|^2,$$ where $X= \pi_\ast E, V= \mathcal{V}E$ and $i_X : \Lambda^2TM \to TM$ is the interior product.
Proof
The ricci tensor of the twistor space is given by $$c_Z(E,E) = c_M (X,X) + t Trace (A \to (\nabla_A R)(\sigma \times V, X)) + (t^2/4)\| \mathcal{R}(\sigma \times V)\|^2 -(t/2) \| i_X \circ \mathcal{R}_+\|^2 + (t/2) \| i_X \circ \mathcal{R}(\sigma)\|^2 + \|V\|^2.$$
Since the manifold M is Hermitian, anti self dual and Einstein, $$\mathcal{J}V = \sigma \times VE,$$ $$\mathcal{R}= (s/6) Id + \mathcal{W}_-, \mathcal{R}_+ = (s/6)Id,$$ and $$g((\nabla_Y R)(W,X),Y)=g((\nabla_Y R)(X \wedge Y),W)=0, $$ for $X,Y \in \mathcal{X}$ and $W \in \mathcal{V}.$ Thus, $$c_Z(E,E) = \sum_{i=1}^4 g(R(X,E_i)X,E_i) -(t/2) \| i_X \circ \mathcal{R}_+\|^2 + (t/2) \| i_X \circ \mathcal{R}(\sigma)\|^2 +(1+ (ts/12)^2)\|V\|^2,$$ where $\{E_1,E_2,E_3,E_4\}$ are orthonormal basis of TM.
Now I'm stuck how to proceed further.Also, $\| i_X \circ \mathcal{R}_+\|^2 - \| i_X \circ \mathcal{R}(\sigma)\|^2 = \sum_{i=1}^4 (g(R(U)X,E_i))^2 + (g(R(\sigma \times U)X,E_i))^2.$
