An electron in an excited state of Li2+ ion has angular momentum 3h/2π. The de Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is:

Angular momentum, L = nħ/2π = 3h/2π ⇒ n = 3
Also, Wavelength λ = h/p = h/mv = h√(r/3h/2π) = 2πr/3 .. (1)
But, r = a0n²/Z (2)
Substituting (2) in (1), λ = 2π(3a0n²/Z)/3 = 2π(3a0(3)²/3) = 2πa0
On comparing with the given expression pπa0 ⇒ p = 2