Given:

$x_m=1.5\times {10}^{-2}$

$T_1=\left(173 °\mathrm{C}+273\right)=100 \mathrm{K}$

From Curie's Law,

$x_m\propto \frac{1}{T}$

$\frac{x_{m_1}}{x_{m_2}}=\frac{T_2}{T_1}$

Substitute the values in the above equation

$\frac{1.5\times {10}^{-2}}{0.5\times {10}^{-2}}=\frac{T_2}{100}$

$T_2=300 \mathrm{K}$

$T_2=300 \mathrm{K}-273$

$T_2=27 °\mathrm{C}$

The change in temperature is,

$\mathrm{\Delta }T=T_1-T$

$=27 °\mathrm{C}-\left(-173 °\mathrm{C}\right)$

$=200 °\mathrm{C}$