Answer
$8.57\;\rm J$
Work Step by Step
First, we need to find $W_{out}$ in the Carnot engine which is equal to $W_{in}$ in the refrigerator.
Both of them, the refrigerator and the heat engines, are having the same reservoirs.
For Carnot heat engines, we know that
$$\eta_1=1-\dfrac{T_C}{T_H}= \dfrac{W_{out}}{Q_{H1}}$$
where 1 refers to the heat engine and 2 refers to the refrigerator.
Hence,
$$W_{out}=Q_{H1}\left[ 1-\dfrac{T_C}{T_H} \right]$$
As we mentioned above, we know that $W_{in}=W_{out}$, so
$$W_{in}=Q_{H1}\left[ 1-\dfrac{T_C}{T_H} \right]\tag 1$$
We also know that the coefficient of performance of the refrigerator is given by
$$K=\dfrac{Q_{C2}}{W_{in}}$$
Hence,
$$Q_{C2}=KW_{in}\tag 2$$
We also know that
$$Q_{H2}=W_{in}+Q_{C2}$$
Plugging from (2);
$$Q_{H2}=W_{in}+KW_{in}=W_{in}(K+1)$$
Plugging from (1);
$$Q_{H2} =Q_{H1}\left[ 1-\dfrac{T_C}{T_H} \right](K+1)$$
Plugging the known;
$$Q_{H2} =(10)\left[ 1-\dfrac{250}{350} \right](2+1)$$
$$Q_{H2} =\color{red}{\bf 8.57}\;\rm J$$
