Here, the diameter of the input piston,$d = 10 \mathrm{cm} = 0.1 \mathrm{m}$

The diameter of output piston, $D = 30 \mathrm{cm} = 0.3 \mathrm{m}$

The weight lifted by the output piston, $F=1.44\times {10}^4 \mathrm{N}$

Let a and A be the areas of a cross-section of the input and output pistons respectively. If $f$ is the force applied on the input piston, then

$\frac{F}{A}=\frac{f}{a}$ or $\frac{F}{\frac{\pi D^2}{4}}=\frac{f}{\frac{\pi d^2}{4}}$

Or $f=F\times \frac{d^2}{D^2}=1.44\times {10}^4\times \frac{{0.1}^2}{{0.3}^2}=1.6\times {10}^3\mathrm{ }\mathrm{N}$