It takes 80 cal/g to change solid water at 0°C to liquid water. How many calories will it take to change 3 g of water at 0 °C to liquid water?

The idea here is that the problem is providing you with a measure of how much heat is needed per gram of solid water, i.e. of ice, in order for a solid #-># liquid phase change to take place.

Simply put, you are given a measure of how much heat is required to melt #"1 g"# of ice at its normal melting point of #0^@"C"#.

So, you can say that #"80 cal g"^(-1)# is equivalent to saying that it takes #"80 cal"# of heat for every #"1 g"# of ice to convert it from solid at #0^@"C"# to lqiuid water at #0^@"C"#.

This means that #"3 g"# of ice would require

#3 color(red)(cancel(color(black)("g"))) * "80 cal"/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("240 cal")))#

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the mass of ice.

So remember this notation

#color(blue)("80 cal")color(red)("/g") " "=" " color(blue)("80 cal")color(white)(.)color(red)("g"^(-1))#

means that every #color(red)("1 g")# of ice at #0^@"C"# can be converted to liquid water at #0^@"C"# by providing it with #color(blue)("80 cal")# of heat.