Answer
a) $303\;\rm kg/year$
b) $1782\;\rm m^2$
Work Step by Step
a)
We know that
$$e=\dfrac{W}{Q_h}$$
whereas $Q_h$ is the total energy used through the year and $W$ is the work done by the refrigerator through the same year.
Solving for $Q_h$;
$$Q_h=\dfrac{W}{e}=\dfrac{2\times10^9}{0.33}=\bf 6.06\times10^9\;\rm J$$
This is the whole energy used throughout the year and to find the amount of burned coal we need to divide this result over the energy that comes from burning 1 kg of coal.
Thus,
$$m_{coal}=\dfrac{Q_h}{E_{coal}}=\dfrac{6.06\times10^9}{2\times10^7}$$
$$m_{coal}=\color{red}{\bf 303}\;\rm kg$$
This is the amount burned in one year.
b)
Since the 10,000 $\rm m^2$ of forest can capture about 1700 kg of $\rm CO_2$ in one year, so the area of forest needed to capture the 303 kg of $\rm CO_2$ is given by
$$A=303\cdot \dfrac{10000}{1700}=\color{red}{\bf 1782}\;\rm m^2$$
