An electricity and magnetism problem by Naren Bhandari

The volume of the wire stays the same. If $A_0$ is the initial cross-sectional area and $L_0$ is the initial length.

$V_0 = A_0 L_0$

If the wire length increases by 50%, the area must decrease by the same factor.

$V = AL = \frac{A_0}{1.5} (1.5 L_0)$

The new resistance is therefore:

$R = \frac{\rho L}{A} = \frac{\rho (1.5 L_0)}{\frac{A_0}{1.5}} = 2.25 R_0$

This is an increase of 125%