$m=300 \,g$

Calorific value $=30 \,kJ / g$

Efficiency, $\eta=10 \%=0.1$

$T_{1}=0^{\circ} C =273 \,K$

$T_{2}=60^{\circ} C =333 \,K$

$e=1$ (black body)

From Newton's law of cooling,

$\frac{d Q}{d t}=e A \sigma\left(T_{2}^{4}-T_{1}^{4}\right)$

$=1 \times 7 \times 10^{2} \times 5.735 \times 10^{8}$

$=26.75 \,W$

$\text { Total heat produced, }$

$H =\eta m \times \text { calorific value }$

$\left.=0.1 \times 300 \times 30 \times 10^{3}-(273)^{4}\right]$

$=9 \times 10^{5} \,J$

$\therefore \text { Time, }$ $t =\frac{\text { Heat produced }(H)}{\text { Rate of emission }\left(\frac{d Q}{d t}\right)}$

$=\frac{9 \times 10^{5}}{26.75 \times 3600}=9.35 \,h \simeq 10 \,h$