Hydraulic lifts work on Pascal's principle, that a change in pressure in what part of an ideal contained fluid is instantly transmitted to all other parts (at the speed of sound in the fluid, which is effectively instantaneous for most purposes where this is used). This means the pressure throughout the fluid is always the same, and since pressure is force per unit area (P = F/A), this means:
(F1/A1) = (F2/A2)
at any points "1" and "2" in the fluid. For a hydraulic lift like this, we often look at the "input" and the "output." In this case, the "input" is where we put in the force to lift, and the "output" is where the output force is applied to whatever we are trying to lift.
In this case, we have:
(Fin/Ain) = (Fout/Aout)
which can be quickly rearranged to:
Fout = (Aout/Ain)(Fin)
This shows the utility of the lift. The output force is greater than the input force by a factor of Aout/Ain (presumimng Aout>Ain, which it has to be for the lift to be useful). This quantity by which the output force is increased in a machine like this is called the "mechanical advantage."
So, you need to have an output force that matches the weight of the car, which is given. You are also given Aout and Ain. So you can use the previous equation to solve immediately for the input force, Fin, needed to hold the car up.
I hope this helps! Just let me know if you want to check and answer or have any more questions about this.