An iron ball is $0.01\mathrm{mm}$ larger to pass through a hole of diameter $6\mathrm{cm}$ in a brass plate when the ball and plate are at a temperature of ${20}^{\circ }\mathrm{C}$  . The temperature in ${ }^0\mathrm{C}$  at which (both for ball and plate) the ball just passes through the hole is (round off to nearest integer) ______ $\left({\mathrm{\alpha }}_{\mathrm{iron}}=12\times {10}^{-6}/{}^0\mathrm{C},{\mathrm{\alpha }}_{\mathrm{brass}}=18\times {10}^{-6}/{}^0\mathrm{C}\right)$