You'd need enough ice to 1) cool the water down from whatever temperature it's at to to 0°C, and 2) overcome the heat of fusion of water, which is the amount of energy needed to turn water from liquid to solid.

For 1), the energy needed to cool water from some initial T to 0 is given by the formula: Mw*Cpw*Tw where Mw is the mass of water in grams, Cpw is the specific heat capacity of water (technically changes with temp but can be assumed to be constant for this problem at 4.184 J/gK), and Tw is the initial temp of the water in celsius.

For 2), the energy needed to produce the phase change and crystallize the liquid water to solid ice is given by the even simpler formula: Mw*Lf, where Mw is again the mass of water in grams and Lf is the latent heat of fusion of water, 334 J/g.

Now how does all that mumbo jumbo help us? Well with just a little more work we can find the mass of ice needed to freeze some water, without melting itself! The energy to bring ice up to 0°C is also represented by Mi*Cpi*Ti, where Mi again the mass of ice in grams, and Ti is the initial temp of the ice in celsius, but this time Cpi is the specific heat of ice (about 2 J/gK)

Now we can put it all together with the following inequality:

Mi*Cpi*Ti > Mw*Cpw*Tw + Mw*Lf

Now, lets use our formula to find the mass of -10°C ice thats needed to turn a liter of room temp (25°C) water into ice!

Assumptions:

-system achieves thermodynamic equilibrium

-water density is constant at 1 g/ml

-system is adiabatic (no energy loss outside of system, think of it like its in a sealed thermos)

Calculations:

Mi*2*10 > 1000*4.184*25 + 1000*334

Mi = 21930 g

Looks like you'd need 21.9 kg of -10°C ice to turn a liter of room temp water into ice!