= Solution
For a single carrier species with positive <charge carrier mobility> $\mu$, the <Hall coefficient> and longitudinal <electrical conductivity> are
$$
R_H=\frac1{nq},\qquad \sigma=n|q|\mu.
$$
Take the <electric current density> along $x$, magnetic field along $z$, and no transverse current. Then $E_y=R_Hj_xB$ and $j_x=\sigma E_x$, giving the <Hall angle> relation
$$
\boxed{\frac{E_y}{E_x}=R_H\sigma B=\operatorname{sgn}(q)\mu B=-\mu B\quad(q=-e).}
$$
The sign specifies the electron Hall polarity; the ratio of field magnitudes is $|E_H|/|E|=\mu|B|$. Converting the stated <electron mobility> to <SI> gives $\mu=900\times10^{-4}=0.0900\,\mathrm{m^2\,V^{-1}\,s^{-1}}$. Therefore
$$
\boxed{|B|=\frac{0.015}{0.0900}=0.167\,\mathrm T.}
$$
This uses the homogeneous single-carrier <Drude model>; the carrier density cancels.
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