= Solution
For one isotropic carrier species in the <Drude model>, the longitudinal <electrical resistivity> at zero transverse <electric current> is $\rho_{xx}=1/(ne\mu)$, and the Hall <electrical resistivity> has magnitude $|\rho_{yx}|=B/(ne)$. Equivalently, the <electric field> balances the magnetic deflection through $E_y=v_xB$, while $|v_x|=\mu|E_x|$. Therefore the magnitude of the <Hall angle> obeys
$$
\frac{|E_y|}{|E_x|}=\mu B.
$$
The required ratio gives
$$
\boxed{B=\frac{0.01}{0.2\,\mathrm{m^2\,V^{-1}\,s^{-1}}}=0.050\,\mathrm T=50\,\mathrm{mT}.}
$$
The <electron density> cancels. With <electric current> along $+x$ and field along $+z$, <Electrons> have negative <Hall coefficient> and hence $E_y<0$; the requested one-percent condition concerns the magnitude. This calculation assumes <Electrons> dominate transport and the transverse contacts draw no <electric current>, so a Hall field can develop.
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