The old convolution trick works for this. (Possibly what Robert Israel had in mind in his answer). Let $\chi_A(x)$ and $\chi_{-B}(x)$ denote the characteristic functions of $A$ and $\{-x: x \in B\}$ respectively, extended to all of ${\mathbb R} $ by setting them equal to zero outside of $[0,1]$. Then $\int_{\mathbb R} \chi_{-B} \ast \chi_A(t) \,dt = \int_{\mathbb R} \int_{\mathbb R} \chi_{-B}(t- x)\chi_A(x)\,dx\,dt$ $= \int_{\mathbb R} \int_{\mathbb R} \chi_{B}(x-t)\chi_A(x)\,dx\,dt$ $ = \int_{\mathbb R} m(A \cap B_t)\,dt$ But by Fubini's theorem the double integral is also $m(A)m(-B) = {1 \over 16}$. Since the $t$-integrand is nonzero only on $[-1,1]$, the integrand must be at least ${1 \over 32}$ for some $t$. Hence for that $t$, one has $m(A \cap B_t) \geq {1 \over 32} > {1 \over 1000}$.