Let $R$ be a ring with identity and $M$ be a injective $R$-module. Is $M^{(N)}$ (a direct sum of copies of $M$) a quasi-injective $R$-module?
Here is my proof.
Since M is injective, M is M-injective hence $M^{(N)}$ is $M$-injective. Therefore $M^{(N)}$ is $M^{(N)}$-injective thus quasiinjective.
Consider the definition that an $R$-module $M$ is injective if the (contra-variant) functor $K\mapsto \operatorname{Hom}(K,M)$ is exact. This means if $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ is an exact sequence, then $$0\rightarrow \operatorname{Hom}(C,M)\rightarrow \operatorname{Hom}(B,M)\rightarrow \operatorname{Hom}(A,M)\rightarrow 0$$ is also an exact sequence.
Then it follows immediately that the sequence $$0\rightarrow \operatorname{Hom}(C,M^{\oplus n})\rightarrow \operatorname{Hom}(B,M^{\oplus n})\rightarrow \operatorname{Hom}(A,M^{\oplus n})\rightarrow 0$$ is exact, since $\operatorname{Hom}(A,M^{\oplus n})\cong\operatorname{Hom}(N,M)^{\oplus n}$ for any $R$-module $N$.
This shows that $M^{\oplus n}$ is injective, not only quasi-injective.
Hope this helps.