There are two heat transfers involved.
#"heat gained by the ice + heat lost by the drink = 0"#
#color(white)(mmmmmm)q_1color(white)(mmmll)+color(white)(mmmmm) q_2color(white)(mmmml) = 0#
#color(white)(mmmm)m_1Δ_text(fus)H color(white)(mml)+ color(white)(mmml)m_2CΔT color(white)(mmll)= 0#
Let's calculate these heats separately.
#m_1color(white)(mll) = ?#
#Δ_text(fus)H = "0.33 kJ/g"#
∴ #q_1 = m_1 × "0.33 kJ/g" = 0.33m_1 color(white)(l)"kJ/g"#
I assume that the drink has the same density as water. Then
#m_2 = 12.0 color(red)(cancel(color(black)("oz"))) × (29.57 color(red)(cancel(color(black)("mL"))))/(1 color(red)(cancel(color(black)("oz")))) × "1.00 g"/(1 color(red)(cancel(color(black)("mL")))) = "354.8 g"#
#ΔT = T_2 - T_1 = "(35 - 75) °F" = "-40" color(red)(cancel(color(black)("°F"))) × "5 °C"/(9 color(red)(cancel(color(black)("°F")))) = "-22.2 °C"#
#q_2 = 354.8 color(red)(cancel(color(black)("g"))) × 4.18 color(white)(l)"J"·color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × "-22.2" color(red)(cancel(color(black)("°C"))) = "-32 920 J" = "-32.92 kJ"#
Now, we can add the two heats.
#q_1 + q_2 = 0.33m_1 color(red)(cancel(color(black)("kJ")))·"g"^"-1" - 32.92 color(red)(cancel(color(black)("kJ"))) = 0#
#m_1 = 32.92/("0.33 g"^"-1") = "100. g"#
