An icebox almost entirely filled with ice at 0^ ^ ^ C is submerged...

The rate of heat transfer through the walls of the icebox is given by Fourier's law of heat conduction:

$\dfrac{dQ}{dt} = \dfrac{kA(T_2 - T_1)}{d}$

Substituting the given values:
k = 0.06 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
$A = 2400 \text{ cm}^2 = 0.24 \text{ m}^2$
T_2 - T_1 = 20 - 0 = 20^{}^{{}^{\circ}}\text{C}
$d = 2.0 \text{ mm} = 2 \times 10^{-3} \text{ m}$

$\dfrac{dQ}{dt} = \dfrac{0.06 \times 0.24 \times 20}{2 \times 10^{-3}} = 144 \text{ J s}^{-1}$

The rate at which the ice melts is given by:

$\dfrac{dm}{dt} = \dfrac{1}{L} \dfrac{dQ}{dt}$

Substituting $L = 3.4 \times 10^5 \text{ J kg}^{-1}$:

$\dfrac{dm}{dt} = \dfrac{144}{3.4 \times 10^5} \text{ kg s}^{-1}$

To find the rate in $\text{kg h}^{-1}$, multiply by $3600 \text{ s}$:

$\dfrac{dm}{dt} = \dfrac{144 \times 3600}{3.4 \times 10^5} \text{ kg h}^{-1}$

$\dfrac{dm}{dt} = \dfrac{518400}{340000} \approx 1.52 \text{ kg h}^{-1}$

Rounding off to two significant figures, the rate at which the ice melts is $1.5 \text{ kg h}^{-1}$.