Answer
${\bf 81}\;\rm mT$
Work Step by Step
We know that Hall voltage is given by
$$\Delta V_{\rm H}=\dfrac{IB}{tne}$$
Let's assume that the only changed thing in the second conductor is the magnetic filed strength that causes a change in the Hall voltage.
In other words, the two conductors are having the same $t$, $n$, $e$, and $I$.
Thus,
$$\dfrac{(\Delta V_{\rm H})_1}{(\Delta V_{\rm H})_2}=\dfrac{B_1}{B_2}$$
Hence,
$$B_2=B_1\left[\dfrac{(\Delta V_{\rm H})_2}{(\Delta V_{\rm H})_1}\right]$$
Plug the known;
$$B_2=(55\rm\; m)\left[\dfrac{2.8\mu}{1.9\mu}\right]$$
$$B_2=\color{red}{\bf 81}\;\rm mT$$
