If there is an algorithm $A$ that guesses the entire message, given an encryption, with probability of $0.2$, then it's not $(O(A) +O(n), 0.1)$-secure
Let $m_1$ be an arbitrary message. We are guaranteed that $P(A(E(k,m_1)) = m_1) = 0.2$. Averaging over all other messages, there is some message $m_2$ such that $P(A(E(k,m_1)) = m_2) \leq 0.8/(2^n-1) \leq 2^{-n}$.
Consider the following algorithm $B$: on input $x$, if $A(x) = m_2$ then output $1$, else output $0$. By assumption, $P(B(E(k,m_1))=1) \leq 2^{-n}$, whereas $P(B(E(k,m_2))=1) = 0.2$. When $n \geq 4$, the difference between the two probabilities is larger than $0.1$.