Answer
The resistance should be made $\frac{1}{3}$ of initial resistance.
Work Step by Step
Let the initial heat be ${{H}_{1}}$ and the final heat be ${{H}_{2}}$.
${{H}_{1}}=k\frac{{{v}^{2}}}{{{r}_{1}}}$ and ${{H}_{2}}=k\frac{{{v}^{2}}}{{{r}_{2}}}$
As, ${{H}_{2}}=3{{H}_{1}}$
$\begin{align}
& 3{{H}_{1}}=k\frac{{{v}^{2}}}{{{r}_{2}}} \\
& 3\frac{k{{v}^{2}}}{{{r}_{1}}}=\frac{k{{v}^{2}}}{{{r}_{2}}} \\
\end{align}$
Solving further,
$\begin{align}
& \frac{1}{3}=\frac{{{r}_{2}}}{{{r}_{1}}} \\
& {{r}_{2}}=\frac{{{r}_{1}}}{3} \\
\end{align}$
That is, to generate the final heat three times of the initial heat, it is required to reduce resistance by one third of initial resistance, keeping voltage constant.
