Since heat is absorbed here at constant pressure so ΔH = 1440 calories
ΔU can be calculated by using the formula :ΔU=ΔH-PΔV
P= 1atm, ΔV = 0.0196L - 0.180L = 0.0016 L
PΔV = 1*0.0016= 0.0016Latm = 0.0016*24.217calories. (1Latm =24.217 calories)
So ΔU = 1440 cal-0.0016*24.217cal≈1439.96 calories .There is negligible work done here.
so you can write ΔH≈ΔU