To solve this problem we will apply Pascal's principle for which we know that the pressure relation between two surfaces under the same fluid is given by the relation between the applied force and the area which receives-prints the force. Mathematically this is,
[tex]P_1 = P_2[/tex]
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
[tex]\frac{F_1}{\pi r_1^2}=\frac{F_2}{\pi r_2^2}[/tex]
For the relation given we have,
[tex]r_1 = 4.8cm[/tex]
[tex]F_2 = 6F_1[/tex]
Replacing,
[tex]\frac{F_1}{\pi 4.8^2}=\frac{6F_1}{\pi r_2^2}[/tex]
[tex]r_2 = \sqrt{(6)(4.8^2)}[/tex]
[tex]r_2 = 11.75cm[/tex]
Finally we have that the work done on both sides should be equal then
[tex]F_1d_1 = F_2d_2[/tex]
[tex]F_1d_1 = 11F_1d_2[/tex]
[tex]d_2 = \frac{x_1}{11}[/tex]
[tex]x_2 = \frac{19.9}{6}[/tex]
[tex]x_2 = 3.31cm[/tex]
