$F=- \eta A \large\frac{dv}{dx}$

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.