1. Use Bohr quantization of angular momentum

For a hydrogen-like ion, $L = n\frac{h}{2\pi} = n\hbar$

Given: $L = \frac{3h}{2\pi}$ So, $n = 3$

Thus the electron is in the $n=3$ state of $\mathrm{Li^{2+}}$.

2. Relate de Broglie wavelength to orbit circumference

For Bohr orbits, $2\pi r_n = n\lambda$ So, $\lambda = \frac{2\pi r_n}{n}$

For a hydrogen-like ion, $r_n = \frac{n^2 a_0}{Z}$ Here for $\mathrm{Li^{2+}}$, $Z=3$. Hence, $r_3 = \frac{3^2 a_0}{3} = 3a_0$

Therefore, $\lambda = \frac{2\pi (3a_0)}{3} = 2\pi a_0$

3. Compare with the given form

Given: $\lambda = p\pi a_0$ From calculation, $\lambda = 2\pi a_0$ So, $p=2$

4. Final answer

The required integer is: $\boxed{2}$

This matches the stored correct answer.