Answer
$V_{4} < V_{1} < V_{3} < V_{2}$
Work Step by Step
$R=\displaystyle \rho(\frac{l}{A}),\qquad V=IR$
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Let the resistance of the cylinder with voltage $V_{3}$ be $R_{3}.$ It has length L and cross section $A=(D/2)^{2}\pi.$
$R_{3}=\displaystyle \rho(\frac{L}{(D/2)^{2}\pi.}),\quad V_{3}=IR_{3}$
$R_{1}=\displaystyle \rho(\frac{3L}{(2D/2)^{2}\pi.})=\frac{3}{4}R_{3},\quad V_{1}=\frac{3}{4}IR_{3}=\frac{3}{4}V_{3}$
$R_{2}=\displaystyle \rho(\frac{2L}{(D/2)^{2}\pi.})=2R_{3},\quad V_{2}=2IR_{3}=2V_{3}$
$R_{4}=\displaystyle \rho(\frac{L}{(2D/2)^{2}\pi.})=\frac{1}{4}R_{3},\quad V_{4}=\frac{1}{4}IR_{3}=\frac{1}{4}V_{3}$
So,
$V_{4} < V_{1} < V_{3} < V_{2}$
