Answer
$(a)\space 2.83\times10^{-5}N$
$(b)\space 10.4\space m/s$
Work Step by Step
(a) Here we use the Stoke's law to find the viscous force,
$F=6\pi\eta Rv$ ; Let's plug known values into this equation.
$F=6\pi(1\times10^{-3}Pa\space s)(5\times10^{-4}m)(3\space m/s)=2.83\times10^{-5}N$
(b) As the sphere reaches its terminal speed,
$F=mg$
$6\pi\eta RV_{t}=mg$
$V_{t}=\frac{mg}{6\pi\eta R}=\frac{(1\times10^{-5}kg)(9.8\space m/s^{2})}{6\pi(1\times10^{-3}Pa\space s)(5\times10^{-4}m)}=10.4\space m/s$
Terminal speed = 10.4 m/s
