Answer
$35.8~grams~$ of ice should be added to the coffee.
Work Step by Step
The heat required to melt the ice and increase in temperature as water is equal in magnitude to the loss of heat from the coffee as it cools:
$m_i~L+m_i~c_w~\Delta T_w = m_c~c_c~\Delta T$
$m_i~(L+c_w~\Delta T_w) = m_c~c_c~\Delta T$
$m_i = \frac{m_c~c_c~\Delta T}{L+c_w~\Delta T_w}$
$m_i = \frac{(0.250~kg)(4186~J/kg~C^{\circ})(20.0~C^{\circ})}{(333\times 10^3~J/kg)+(4186~J/kg~C^{\circ})(60.0~C^{\circ})}$
$m_i = 0.0358~kg$
$m_i = 35.8~g$
$35.8~grams~$ of ice should be added to the coffee.
