1. Concept used

The ice is at $0^\circ\mathrm{C}$ and the surroundings are at $40^\circ\mathrm{C}$. Heat enters through the insulating walls by conduction.

The rate of heat flow is

$\frac{Q}{t}=\frac{kA\Delta T}{x}$

where:

• $k=0.05\,\mathrm{Wm^{-1}\,^{\circ}C^{-1}}$
• $x=1\,\mathrm{cm}=0.01\,\mathrm{m}$
• $\Delta T=40^\circ\mathrm{C}$
• $A$ = total surface area of the ice box

The heat entering is used to melt ice:

$\frac{Q}{t}=\dot m L$

So,

$\dot m=\frac{1}{L}\cdot \frac{kA\Delta T}{x}$

with $L=3.4\times 10^5\,\mathrm{J/kg}$.

2. Dimensions of the box / ice cube

Given dimensions:

$l=60\,\mathrm{cm}=0.6\,\mathrm{m},\quad b=50\,\mathrm{cm}=0.5\,\mathrm{m},\quad h=20\,\mathrm{cm}=0.2\,\mathrm{m}$

Total surface area:

$A=2(lb+bh+hl)$

$A=2\big[(0.6)(0.5)+(0.5)(0.2)+(0.2)(0.6)\big]$

$A=2(0.3+0.1+0.12)=2(0.52)=1.04\,\mathrm{m^2}$

3. Rate of heat conduction

$\frac{Q}{t}=\frac{kA\Delta T}{x}$

$\frac{Q}{t}=\frac{(0.05)(1.04)(40)}{0.01}$

First compute numerator:

$0.05\times 1.04=0.052$

$0.052\times 40=2.08$

Then divide by $0.01$:

$\frac{Q}{t}=208\,\mathrm{W}=208\,\mathrm{J/s}$

4. Convert heat rate into melting rate

$\dot m=\frac{Q/t}{L}=\frac{208}{3.4\times 10^5}$

$\dot m\approx 6.12\times 10^{-4}\,\mathrm{kg/s}$

This can be written as

$\dot m\approx 61\times 10^{-5}\,\mathrm{kg/s}$

5. Match with options

The value matches:

Option B: $61\times 10^{-5}\,\mathrm{kg\,s^{-1}}$

Note: Option C ($208\,\mathrm{kg/s}$) is actually the heat flow in watts, not the mass melting rate.

6. Comparison with stored correct answer

Stored correct answer = B

Derived answer = B

So they agree.