Thickness of the boiler, $l=1.0 cm =0.01 m$

Boiling rate of water, $R=6.0 kg / min$

Mass, $m=6 kg$

Time, $t=1 \min =60 s$

Thermal conductivity of brass, $K=109 Js ^{-1} m ^{-1} K ^{-1}$

Heat of vaporisation, $L=2256 \times 10^{3} J kg ^{-1}$

The amount of heat flowing into water through the brass base of the boiler is given by

$\theta=\frac{K A\left(T_{1}-T_{2}\right) t}{l}\dots (i)$

Where,

$T_{1}=$ Temperature of the flame in contact with the boiler

$T_{2}=$ Boiling point of water $=100^{\circ} C$

Heat required for boiling the water

$\theta=m L \ldots(i i)$

Equating equations $(i)$ and $(i i),$ we get:

$\therefore m L=\frac{K A\left(T_{1}-T_{2}\right) t}{l}$

$T_{1}-T_{2}=\frac{m L l}{K A t}$

$=\frac{6 \times 2256 \times 10^{3} \times 0.01}{109 \times 0.15 \times 60}$

$=137.98^{\circ} C$

Therefore, the temperature of the part of the flame in contact with the boiler is $237.98\,^{\circ} C$