Answer
46.25 grams of aluminum must be dropped in the hole.
Work Step by Step
The heat required to make the ice melt is equal in magnitude to the loss of heat from the aluminum as it cools:
$m_a~c_a~\Delta T = m_i~L$
$m_a = \frac{m_i~L}{c_a~\Delta T}$
$m_a = \frac{(0.010~kg)(333\times 10^3~J/kg)}{(900~J/kg~C^{\circ})(80.0~C^{\circ})}$
$m_a = 0.04625~kg$
$m_a = 46.25~grams$
46.25 grams of aluminum must be dropped in the hole.
