Answer
$\approx 21\%$
Work Step by Step
We assume that the system, the iron block plus the water, is an isolated system.
Thus,
$$Q_{\rm Iron}+Q_{\rm water} =0$$
The gained energy by the water will heat it up to 100$^\circ$C and then the rest of the energy will evaporate part of it which we call the heat of evaporation.
Thus,
$$m_{\rm Iron}c_{\rm Iron}\Delta T_{\rm Iron}+(m_{\rm water}c_{\rm water}\Delta T_{\rm water}+m_{1}L_{\rm v,water}) =0$$
where $m_1$ is the mass of the evaporated water.
We know that the final temperature of the two materials will be the same which is 100$^\circ$C since this is the maximum temperature could water reach without evaporating.
Thus,
$$m_{\rm Iron}c_{\rm Iron} (T_f- T_{\rm Iron})+ m_{\rm water}c_{\rm water}(T_f- T_{\rm water})+\\m_{1}L_{\rm v,water} =0$$
Solving for $m_1$;
$$m_1=\dfrac{m_{\rm Iron}c_{\rm Iron} (T_f- T_{\rm Iron})+ m_{\rm water}c_{\rm water}(T_f- T_{\rm water})}{-L_{\rm v,water} }$$
We know that the mass of the iron box and the mass are given by the density law; $m=\rho V$
$$m_1=\dfrac{\rho_{\rm Iron}V_{\rm Iron}c_{\rm Iron} (T_f- T_{\rm Iron})+ \rho_{\rm water}V_{\rm water}c_{\rm water}(T_f- T_{\rm water})}{-L_{\rm v,water} }$$
Plugging the known;
$$m_1=\dfrac{(7870)(65\times 10^{-6}) (449) (100- 800)+(1000) (0.2\times 10^{-3})(4190)(100- 20)}{-(22.6\times 10^5) }$$
$$m_1=\bf \color{red}{\bf 0.0415}\;\rm kg=\bf 41.5\;\rm g$$
The fraction of water that boils away is given by
$$f=\dfrac{m_1}{m_{\rm water}}\times100\%$$
$$f=\dfrac{m_1}{\rho_{\rm water}V_{\rm water}}\times100\%$$
Plugging the known;
$$f=\dfrac{0.0415}{ (1000)(0.2\times 10^{-3})}\times100\%= 20.75\%$$
$$f\approx \color{red}{\bf 21}\%$$
