Now velocity of water coming out of the hole $=v=\sqrt {2gh}$
$\qquad= {2 \times 10 \times 0.6}$
$\qquad= \sqrt {12}\;m/s$
If a is area of cross section of the hole.
$a \times v$= volume of water flowing per second.
$\therefore\; Thrust =\large\frac{Volume}{Second} $$ \times \rho \times velocity$
$Thrust = \rho av^2$
$\qquad= 10^3 \times \large\frac{\pi}{4}$$ (10^{-2})^2 (\sqrt {12})^2$
$\qquad= \large\frac{3.14 \times 10^{-1} \times 12}{4}$
$\qquad= 0.942\;N$
Hence b is the correct answer.