= Solution
Write $g=(g_1,\ldots,g_n)$ for the continuous representatives of the <weak derivatives>. Integrating the preceding identity over $B=B(0,1)$ gives
$$
|B|u_\varepsilon(x)=\int_Bu_\varepsilon(y)\,dy+\int_B\int_0^1(\rho_\varepsilon*g)(y+t(x-y))\cdot(x-y)\,dt\,dy.
$$
The first term converges to $\int_Bu$ because $u_\varepsilon\to u$ in $L^1$. For fixed $x$, all segments in the second term lie in a compact set. Continuity of $g$ implies $\rho_\varepsilon*g\to g$ uniformly on that set. Hence every $x$ has the finite limit
$$
U(x)=\frac1{|B|}\left(\int_Bu(y)\,dy+\int_B\int_0^1g(y+t(x-y))\cdot(x-y)\,dt\,dy\right).
$$
The same argument is uniform for $x$ in any fixed compact set: all the segments then lie in a common compact set. Thus $u_\varepsilon\to U$ locally uniformly, so $U$ is continuous; the local $L^1$ convergence also gives $U=u$ <almost everywhere>.
Passing to the limit in the segment identity, and taking its endpoints to be $x$ and $x+h$, gives
$$
U(x+h)-U(x)-g(x)\cdot h=\int_0^1[g(x+th)-g(x)]\cdot h\,dt=o(|h|).
$$
Therefore $U$ is <differentiable> with continuous <gradient> $g$: $\boxed{U\in C^1,\quad U=u\text{ almost everywhere},\quad\nabla U=g}$. This is the <continuous weak gradient gives a C1 representative> criterion.
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