Let $(X, \mathcal B, \mu)$ be a "good" measure space, e.g. $\mu$ is a positive Radon measure on a locally compact topological space $X$ with Borel $\sigma$-algebra $\mathcal B$. Let $A\subset X$ such that every measurable subset of $A$ has zero measure. Is it true that there is a zero measure set $B$ such that $A\subset B$?
You are asking whether every set with inner measure $0$ has measure $0$ with respect to the completion measure. The Lebesgue measure, for example, does not have this property, since the usual Vitali set is non-measurable, but has inner measure $0$.
I claim that a ($\sigma$-finite) measure space has your property if and only if every set is measurable with respect to its completion measure.
For the forward direction, suppose that $A$ is any subset of the space $X$. Let $a$ be the inner measure of $A$, the supremum of $\mu(A_0)$ among all measurable $A_0\subset A$ (and we may assume wlog this is finite). By taking a union, it follows that the inner measure is realized, so that there is some measurable $A_0\subset A$ with $\mu(A_0)=a$. It follows that $A-A_0$ has inner measure $0$. By the hypothesis, it follows that $A-A_0\subset B$ for some measurable set $B$ with $\mu(B)=0$, and so $A-A_0$ has measure $0$ with respect to the completion. Consequently, $A$ differs from the measurable set $A_0$ on a completion-measure zero set $A-A_0$, and hence is measurable with respect to the completion measure.
Conversely, suppose that every set is measurable with respect to the completion of $\mu$. Suppose that $A$ has inner measure $0$. By assumption, there is some measurable set $A_0$ such that the symmetric difference $A\triangle A_0\subset A_1$ for some measurable $A_1$ with $\mu(A_1)=0$. It follows that $A_0-A_1$ is a measurable subset of $A$, and hence measure $0$, and so $A\subset B=A_0\cup A_1=(A_0-A_1)\cup A_1$ shows that $A$ is contained in a measure $0$ set $B$, as desired.
In particular, a complete measure has the property if and only if it measures every set.