= Solution
Neglect the battery's internal resistance, contact drops and the very small resistance of laboratory-length metal leads. The germanium resistance and field are
$$
R_g=\rho_g\frac{L_g}{A_g}=0.5\frac{0.01}{10^{-6}}=5000\,\Omega,
\qquad E_g=\frac2{0.01}=200\,\mathrm{V\,m^{-1}}.
$$
Taking the positive direction along the <electric field>, the carrier drift velocities are
$$
\boxed{\mathbf v_e=-72\,\hat{\mathbf E}\,\mathrm{m\,s^{-1}},\qquad
\mathbf v_h=+36\,\hat{\mathbf E}\,\mathrm{m\,s^{-1}}.}
$$
Their conventional current densities add: $j_e=en_i\mu_e E_g\simeq266.7\,\mathrm{A\,m^{-2}}$ and $j_h\simeq133.3\,\mathrm{A\,m^{-2}}$. Thus $I=(j_e+j_h)A_g=0.400\,\mathrm{mA}$ is the same total current through either metal lead. The <current continuity between a metal and an intrinsic semiconductor> gives
$$
\boxed{|v_m|=\frac{I}{n_meA_m}
=\frac{4.00\times10^{-4}}{(8.57\times10^{28})e(0.5\times10^{-6})}
=5.82\times10^{-8}\,\mathrm{m\,s^{-1}}.}
$$
The metal <electrons> move opposite its conventional current. This extremely small drift speed is compatible with large microscopic Fermi velocities; drift is the small imbalance of the occupied <momentum> distribution.
The wire lengths were not specified. Their resistance per metre is $\rho_m/A_m=0.034\,\Omega\,\mathrm{m^{-1}}$, so neglecting them relative to $5000\,\Omega$ is justified for an ordinary apparatus. If finite total wire length $L_w$ is retained, the exact ideal-contact circuit expression is $I=2/(5000+0.034L_w)\,\mathrm A$ with $L_w$ in metres; all three drift magnitudes above scale with $I/(0.400\,\mathrm{mA})$. Thus the approximation and its missing geometric input are explicit.
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