Answer
$ 1.31\times 10^{-4}\;\rm F$
Work Step by Step
First of all, we need to find the work done on the block to lift it up a distance of 3 m. This work is equal to the change in gravitational potential energy.
$$W=\Delta U_g $$
$$W=mgh $$
Also, this work is done by 90% of the output of the motor.
Thus
$$0.9\Delta U_C=mgh $$
where the motor is using a fully charged capacitor.
$$0.9\left(\frac{1}{2}CV_C^2\right)=mgh $$
Solving for $C$;
$$ C=\dfrac{2mgh}{0.9V_C^2} $$
Plug the known;
$$ C=\dfrac{2(2)(9.8)(3)}{0.9(1000)^2} $$
$$C=\color{red}{\bf 1.31\times 10^{-4}}\;\rm F$$
