Difference in the pressure at the two positions is equal to the pressure due to liquid column of AB.
Pressure at larger position $P_1=\large\frac{F}{A_1}$
$\qquad= \large\frac{F}{\pi \bigg(\Large\frac{0.35}{2}\bigg)^2}$
Pressure at smaller position $P_2=\large\frac{20}{A_2}$
$\qquad= \large\frac{20 \times g}{\pi \bigg(\Large\frac{0.1}{2}\bigg)^2}$
$P_2-P_1=h \rho g$
$\large\frac{20 \times 9.8 }{\pi \times (5 \times 10^{-2})^2}- \large\frac{F}{\pi \times (17.5 \times 10^{-2})^2}$$= 1.5 \times 750 \times 9.8$
Solving we get,
$F= 1.3 \times 10^3\;N$
Hence c is the correct answer