Water, with η = 1.002 mPa s, flows through a horizontal capillary tube of radius 0.50 mm and length…

Step 1. The flow rate. $Q = \dfrac{\pi \times 2000 \times (5.0 \times 10^{-4})^4}{8 \times 1.002 \times 10^{-3} \times 0.20} = \dfrac{3.927 \times 10^{-10}}{1.603 \times 10^{-3}} = 2.45 \times 10^{-7}\ \text{m}^3/\text{s}$. One $\text{m}^3$ is $10^6\ \text{mL}$, so $Q = 0.245\ \text{mL/s}$.

Step 2. The time. $t = \dfrac{50\ \text{mL}}{0.245\ \text{mL/s}} = 204\ \text{s}$.

Step 3. Half the radius. $Q \propto \Delta P\,r^4$. With $r$ halved, $r^4$ falls to $\tfrac{1}{16}$. To keep $Q$ the same, $\Delta P$ must be $16$ times larger: $16 \times 2.0\ \text{kPa} = 32\ \text{kPa}$.