= Solution
Using metres and SI <viscosity>, the pressure difference is $\Delta P=200$ Pa and the central flow estimate is
$$
Q=\frac{\pi(2.3\times10^{-3})^4(200)}{8(8.9\times10^{-4})(0.50)}=4.939\times10^{-6}\,\mathrm{m^3\,s^{-1}}.
$$
If the quoted uncertainties are independent <standard uncertainties>, the pressure-difference uncertainty is $\sigma_{\Delta P}=\sqrt{10^2+10^2}=14.14$ Pa. First-order propagation gives
$$
\left(\frac{\sigma_Q}{Q}\right)^2=\left(4\frac{0.1}{2.3}\right)^2+\left(\frac{0.3}{8.9}\right)^2+\left(\frac{0.01}{0.50}\right)^2+\frac{10^2+10^2}{200^2}=0.03678.
$$
Hence
$$
\boxed{Q=(4.94\pm0.95)\times10^{-6}\,\mathrm{m^3\,s^{-1}}=(4.94\pm0.95)\,\mathrm{mL\,s^{-1}}.}
$$
The <radius sensitivity of Poiseuille-flow uncertainty> comes from the fourth-power dependence on radius. If the quoted errors instead mean independent worst-case bounds, the first-order absolute fractional bound is $4(0.1/2.3)+0.3/8.9+0.01/0.50+20/200=0.3276$, giving approximately $\pm1.62\times10^{-6}$ m³/s. That conservative bound and the <standard uncertainty> result are different conventions, not interchangeable errors. The supplied flow model presumes a <Newtonian fluid> and <laminar flow> that is fully developed; no <density> is given for an independent Reynolds-number check.
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