A nuclear fuel rod generates heat at a volumetric rate of $4.50 \times 10^4 \text{ Btu}/(\text{ft}^3\cdot\text{h})$. Convert this volumetric heat generation rate to kilowatts per cubic meter ($\text{kW/m}^3$). (Take $1 \text{ Btu} \approx 1055.056 \text{ J}$, $1 \text{ ft} \approx 0.3048 \text{ m}$, and $1 \text{ h} = 3600 \text{ s}$)