Answer
$R = 54~\Omega$
Work Step by Step
Let $L_0$ be the original length.
Let $A_0$ be the original cross-sectional area.
Since the density is the same, the volume remains the same.
We can find the new area:
$A~L = V = A_0~L_0$
$A~(3~L_0) = A_0~L_0$
$A = \frac{A_0}{3}$
We can find an expression for the original resistance:
$R_0 = \frac{\rho~L_0}{A_0}$
We can find the resistance of the longer wire:
$R = \frac{\rho~L}{A}$
$R = \frac{\rho~(3~L_0)}{A_0/3}$
$R = 9~\frac{\rho~L_0}{A_0}$
$R = 9~R_0$
$R = (9)~(6.0~\Omega)$
$R = 54~\Omega$
