Answer
$2.25 \Omega$.
Work Step by Step
The amount of metal is unchanged, so the volume stays the same.
If the length of the wire increases by a factor of 1.5, the cross-sectional area decreases by the same factor of 1.5.
$$R_f=\rho\frac{\mathcal{l}_f}{A_f}=\rho\frac{1.5\mathcal{l}_i}{A_i/1.5}=2.25\rho\frac{\mathcal{l}_i}{A_i}=2.25R_i$$
The final resistance is 2.25 times the old resistance of $1.0\Omega$, so it is $2.25 \Omega$.
