Answer
$189\,\Omega$
Work Step by Step
Let $A_{1}$ be the initial cross-sectional area, $l_{1}$ be the initial length, $A_{2}$ be the final cross-sectional area and $l_{2}$ be the final length.
Given, $l_{2}=3l_{1}$
Since volume is constant,
$A_{1}l_{1}=A_{2}l_{2}\implies A_{2}=\frac{A_{1}l_{1}}{l_{2}}=\frac{A_{1}l_{1}}{3l_{1}}=\frac{A_{1}}{3}$
$R_{1}=\rho \frac{l_{1}}{A_{1}}=21.0\,\Omega$
$R_{2}=\rho \frac{l_{2}}{A_{2}}=\rho \frac{3l_{1}}{\frac{A_{1}}{3}}=9\rho\frac{l_{1}}{A_{1}}=9\times21.0\,\Omega=189\,\Omega$
