There are two heats involved in this problem.

#"heat of solution of KOH + heat absorbed by water" = 0#

#color(white)(mmmmm)q_1color(white)(mmmmmll) +color(white)(mmmmm) q_2color(white)(mmmmml) = 0#

#color(white)(mmmm)nΔHcolor(white)(mmmmmmmml) + mCΔT color(white)(mmmml)= 0#

Let's calculate the heats separately.

#n = 13.90 color(red)(cancel(color(black)("g KOH"))) × "1 mol KOH"/(56.11 color(red)(cancel(color(black)("g KOH")))) = "0.2477 mol KOH"#

#q_1 = nΔH = 0.2477ΔH color(white)(l)"mol"#

#m = "(13.90 + 122.40) g" = "136.30 g"#

#C = "4.184 J·g·°C"^"-1"#

#ΔT = "(45.73 - 20.70) °C" = "25.03 °C"#

#q_2 = mCΔT = 136.30 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g"^"-1"·"°C"^"-1"))) × 25.03 color(red)(cancel(color(black)("°C"))) = "14 274 J" = "14.274 kJ"#

#q_1 + q_2 = 0.2477ΔH color(white)(l)"mol" + "14.274 kJ" = 0#

#ΔH = "-14.274 kJ"/"0.2477 mol" = "-57.62 kJ/mol"#