The answer is yes.

We construct $f$ by transfinite recursion using a well ordering of the reals. (So this may not be explicit enough for you.)

In fact, we can make the function $f$ bijective, with the graph of $f$ having full outer measure in every rectangle.

**Theorem.** There is a bijective function $f:\mathbb{R}\to\mathbb{R}$ with full outer measure in every rectangle in the plane.

**Proof.** There are only continuum many Borel sets, and so we may enumerate a list $(N_\alpha,R_\alpha)$, for $\alpha<\frak{c}$ in type continuum, where $N_\alpha\subset R_\alpha$ is a set with less than full measure in rectangle $R_\alpha$ in the plane, and all such combinations arise in the list.

We now define the function $f$ in stages. At any stage $\alpha<\frak{c}$, we will have defined $f$ on fewer than continuum many points. At stage $\alpha$, we consider the set $N_\alpha\subset R_\alpha$ sitting inside that rectangle. Because it has has less than full measure, the complement $R_\alpha\setminus N_\alpha$ has positive measure, and so there must be continuum many $x$ on whose section in $R_\alpha$ there are points in $R_\alpha\setminus N_\alpha$. So there must some such $x$ on which $f$ is not yet defined, and we may define $f(x)$ so that $(x,f(x))$ is one of those points in $R_\alpha\setminus N_\alpha$. Continue this process for continuum many stages, and then extend $f$ on the remaining points arbitrarily. Since on any rectangle, the graph of $f$ is not contained in any measurable set of less than full measure, the graph must have full outer measure on all such rectangles.

We can arrange that $f$ is onto by changing its values on a size continuum measure zero set, which will not affect the outer measure property of its graph.

We can actually make $f$ bijective, since at each stage in the construction, there will be a not-yet-defined section $x$ on which $N_\alpha$ omits continuum many points in $R_\alpha$, and so at that stage we can let $f(x)$ be a totally new value, while still avoiding $N_\alpha$. And then at the end, we can change $f$ on measure zero set so as to hit all the other values, so $f$ will be bijective. $\Box$

Regarding the question of "explicit", we have seen on the other question that there can be no measurable function with a postive outer measure graph. In particular, there can be no Borel function, and this is a common sense of explicitness. Meanwhile, it is consistent with ZFC that there is a projectively definable well-ordering of the reals, and in this model the function $f$ I provide can be projectively definable, at a fairly low level of complexity. Meanwhile, it is also relatively consistent with ZFC that every projective set is measurable. So the question of whether there is an explicit function $f$ as desired, if this is taken to mean projectively definable, is independent of ZFC.