The Carnot efficiency is
$\begin{align*}e = 1 - \frac{T_{\text{low}}}{T_{\text{high}}}\end{align*}$
where $T_{\text{low}}$ and $T_{\text{high}}$ are the low and high temperatures in $^{\circ} \mbox{ K}$ . So, in this case,
$\begin{align*}e = 1 - \frac{\left(273 +7\right) ^{\circ} \mbox{ K}}{\left(273 + 27\right) ^{\circ} \mbox{ K}} = 1- \frac{280...$
The efficiency is defined as the ratio of the work
$\begin{align*}e = \frac{W}{Q_{\text{in}}}\end{align*}$
where $W$ is the work done by the Carnot engine and $Q_{\text{in}}$ is the heat input to the engine. In this problem, the engine is being run in reverse as a heat pump, so the work $W$ is the work done to the engine and the denominator of $e$ is the heat output by the heat pump.
So, the minimum amount of work required to put out $15000 \mbox{ J}$ of heat is
$\begin{align*}\left(15000 \mbox{ J}\right) \cdot \left(\frac{1}{15} \right) = \boxed{1000 \mbox{ J}}\end{align*}$
Therefore, answer (B) is correct.

Solution to 2008 Problem 35