Solution:
Heat supplied \(\Delta Q=M S \Delta T\)
For same material 's' remains the same
\(\Delta Q \propto M\) and \(M=\frac{4}{3} \pi r^{3} \rho\)
\(\Rightarrow \frac{\Delta Q_{1}}{\Delta Q_{2}}\)
\(=\left(\frac{r_{1}}{r_{2}}\right)^{3}\)
\(=\left(\frac{1.5}{1}\right)^{3}\)
\(=\frac{27}{8}\)