Specific heat of ice in #"J/kg K"#?

Your starting point here will be the specific heat of ice expressed in joules per gram Kelvin, #"J g"^(-1)"K"^(-1)#, which is listed as being equal to

#c_"ice" = "2.06 J g"^(-1)"K"^(-1)#

This tells you that in order to increase the temperature of #"1 g"# of ice by #"1 K"# you must provide it with #"2.06 J"# of heat.

Your goal here is to determine the specific heat of ice in joules per kilogram Kelvin, #"J kg"^(-1)"K"^(-1)#, which essentially tells you how much heat is required in order to increase the temperature of #"1 kg"# of ice by #"1 K"#.

The conversion factor that takes you from grams to kilograms is

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))#

You can use this conversion factor to get

#2.06"J"/(color(red)(cancel(color(black)("g"))) * "K") * (10^3color(red)(cancel(color(black)("g"))))/("1 kg") = color(green)(|bar(ul(color(white)(a/a)color(black)(2.06 * 10^3color(white)(a) "J kg"^(-1)"K"^(-1))color(white)(a/a)|)))#

So, in order to increase the temperature of #"1 kg"# of ice by #"1 K"#, you must provide it with

#2.06 * 10^3"J" = "2060 J"#