What is the Enthalpy of Decomposition of liquid Hydrogen peroxide (in kJ/mol)?

Presumably you want to know:

H2O2(l) → ½ O2(g) + H2O(l), ΔH = ???

I don't see any source for that directly, but the source below gives these (both at 25°C):

H2(g) + O2(g) → H2O2(l), ΔH = -45.16 kcal/mol
H2(g) + ½ O2(g) → H2O(l), ΔH = -68.3174 kcal/mol

So write the first given equation backwards:
H2O2(l) → H2(g) + O2(g), ΔH = +45.16 kcal/mol
Add the second given equation:
H2O2(l) + H2(g) + ½ O2(g) → H2(g) + O2(g) + H2O(l), ΔH = +45.16 - 68.3174 kcal/mol
Cancel like amounts of O2, and all the H2, on both sides of the arrow,
and do the arithmetic for ΔH, then convert to the desired units:
H2O2(l) → ½ O2(g) + H2O(l), ΔH = -23.16 kcal/mol = -96.89 kJ/mol

(If this is for anything besides schoolwork, you would be wise to check it with a second source.)