Here, $K=109 \;\mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}};$ $A=0.15 \;\mathrm{m}^{2} ;\ d=1.0 \;\mathrm{cm}=10^{-2} \;\mathrm{m};$$T_{2}=100^{\circ}\mathrm{C}$$t = 1 \;\mathrm{min} = 60 \;\mathrm{s}$Let $T_1$ be the temperature of the part of boiler in contact with the stove.Therefore, amount of heat flowing per minute through the base of the boiler,$Q= \frac{K A\,(T_{1}-T_{2})}{d}$$= \frac{109 \times 0.15 \times (T_{1}-100) \times 60}{10^{-2}}$Also, heat of vaporisation of water, $L=2256 \times 10^{3} \;\mathrm{J\,kg^{-1}}$Rate of boiling of water in the boiler per minute$M = 6.0 \;\mathrm{kg} = 6000 \;\mathrm{g}$Therefore, heat received by water per minute,$Q = ML = 6000 \times 2256 \;\mathrm{J}$$\frac{109 \times 0.15 \times (T_{1}-100) \times 60}{0.01}$$=6000 \times 2256$$T_{1}-100= \frac{6000 \times 2256 \times 0.01}{109 \times 0.15 \times 60}$$=138^{\circ}\mathrm{C}$or $T_{1}=138+100=238^{\circ}\mathrm{C}$