Side of the given cubical ice box, $s=30 cm =0.3 m$

Thickness of the ice box, $l=5.0 cm =0.05 m$

Mass of ice kept in the ice box, $m=4 kg$

Time gap, $t=6 h =6 \times 60 \times 60 s$

Outside temperature, $T=45^{\circ} C$

Coefficient of thermal conductivity of thermacole, $K=0.01 J s ^{-1} m ^{-1} K ^{-1}$

Heat of fusion of water, $L=335 \times 10^{3} J kg ^{-1}$ Let

$m$ be the total amount of ice that melts in $6 h$.

The amount of heat lost by the food:

$\theta=\frac{K A(T-0) t}{l}$

Where,

$A=$ Surface area of the box $=6 s^{2}=6 \times(0.3)^{2}=0.54 m ^{3}$

$\theta=\frac{0.01 \times 0.54 \times(45) \times 6 \times 60 \times 60}{0.05}=104976 J$

But $\theta=m^{\prime} L$

$\therefore m^{\prime}=\frac{\theta}{L}$

$=\frac{104976}{335 \times 10^{3}}=0.313 kg$

Mass of ice left $=4-0.313=3.687 kg$

Hence, the amount of ice remaining after $6 h$ is $3.687\, kg .$