Let $R$ be a commutative Noetherian ring, $M$ is a finitely generated $R$module. If $F: Mod \to Mod$ is a left exact functor and $R^iF(E)=0$ where $E$ is injective module. Assume that $F() \cong Hom(M,)$, can we infer the $ith$ right devired functors $R^iF()\cong Ext^i(M,)$?

Yes. For example, if you compute right derived functors by injective resolutions, then naturality of the isomorphism between $F$ and $\text{Hom}(M,)$ will ensure that you have an isomorphism between the two complexes whose cohomology groups give you the two derived functors. 

