The poster is using for heat of fusion for water. The transition from one gram of ice at 0°C to one gram of water at 0°C releases 80 calories of heat (or 334 joules). So you need to know roughly how much heat is needed to raise the temperature of one gram of gold by 100°C (which is where my guess of 16 cal fell apart).
Hoodat, Hoodat, Hoodat, what we gonna do wid choo?
Ice melting to water ABSORBS 80 calories per gram, it doesn't "release" it. Confusing words is easy when you know what you mean but misspeak.
Water holds 33 times as much heat as an equivalent weight of gold, and then multiply that by 80 for the heat of fusion.
The answer is 0.617 ounces of ice.
Water has a heat capacity of 4.18 J/g C degree
Gold has a heat capacity of 0.129 J/g C degree
The heat of fusion of water is 80 times the heat capacity of water, so one gram of gold will cool down 100 degrees C with the loss of 12.9 Joules. Ice will melt absorbing 4.18 J x 80 per gram or 334.4 J without changing temperature!
The ratio of 12.9 to 334.4 or .038576 will apply to any respective quantity of gold cooling 100 C and ice just melting without even changing temperature. So a pound of gold at 100 C requires .038576 pound of ice at 0 to cool it. .038576 of 16 ounces is 0.617 ounces of ice.
[Your estimate of .16 specific heat for gold was really good! Only off by a factor of 2.]