$$
p=frac{12200text{ N}}{A_2} = frac{12200text{ N}}{r_2^2 pi} =107573text{ Pa}.
$$

To achieve this pressure under the small piston we need

$$
p=frac{715text{ N}}{r_1^2 pi}
$$
which yields

$$
r_1 = sqrt{frac{715text{ N}}{ppi}} = 4.6text{ cm}.
$$

Since we know that the pressure is the same on both pistons of the lift we can write:

$$
begin{align*}
P_1 &= P_2 \
frac{F_1}{A_1} &= frac{F_2}{A_2}
end{align*}
$$

Now using $F_2 = 72 cdot F_1$ we have:

$$
begin{align*}
frac{A_2}{A_1} = frac{F_2}{F_1} = 72
end{align*}
$$

Now we find the distance piston 2 moved, since we know piston 1 moved for $d_1 = 24text{ cm}$, by equation the volumes of the pushed fluid:

$$
begin{equation*}
A_1 cdot d_1 = A_2 cdot d_2
end{equation*}
$$

Rearranging:

$$
begin{align*}
d_2 = d_1 cdot frac{A_1}{A_2} = d_1 , frac{1}{72}
end{align*}
$$

Plugging in the values we get:

$$
begin{align*}
d_2 = frac{24text{ cm}}{72} = 0.33text{ cm}
end{align*}
$$