Answer
$T=-213^oC$
Work Step by Step
Given:
$l_{bi}=0.04378m$
$l_{ii}=0.04371m$
$T_i=15^oC$
$l_{bi}+\Delta l_{b}=l_{ii}+\Delta l_{i}$
$l_{bi}+\alpha_b l_{ib}\Delta T=l_{ii}+\alpha_i l_{ii}\Delta T$
$\Delta T=\frac{l_{bi}-l_{ii}}{\alpha_i l_{ii}-\alpha_b l_{ib}}=\frac{0.04378m-0.04371m}{(12\times10^{-6})(0.04371m)-(19\times10^{-6})(0.04378m)}=-227.79^oC$
$15^oC-227.79^oC=-213^oC$