$F=- \eta A \large\frac{dv}{dx}$
When $\eta=15.5\;p\qquad A=100\;cm^2$
$\large\frac{dv}{dx}=\frac{3}{0.2}$$=15 s^{-1}$
$F=-15.5 \times 100 \times 15$
$\qquad= -23250\; dynes$
$\qquad= -0.233 N$
force required $=0.233\;N$
Hence b is the correct answer.