$\alpha  = \frac{{{T_2}}}{{{T_1} - {T_2}}}$

where $T_1$ and $T_2$ are the temperatures of hot and cold reservoirs $(in\,kelvin)$ respectively.

$Here,\alpha  = 5,{T_2} =  - {20^ \circ }C =  - 20 + 273\,K = 253\,K$

${T_1} = ?$

$\therefore \,\,5 = \frac{{253K}}{{{T_1} - 253\,K}}$

$5{T_1} - 5\left( {253\,K} \right) = 253\,K$

$5{T_1} = 253\,K + 5\left( {253\,K} \right) = 6\left( {253\,K} \right)$

${T_1} = \frac{6}{5}\left( {253K} \right) = 303.6\,K = 303.6 - 273$

$ = {30.6^ \circ }C \approx {31^ \circ }C$