We know that the de Broglie wavelength is given by:
\[
\lambda = \frac{h}{mv}
\]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.
Let \( v_p \) be the velocity of the proton and \( v_A \) be the velocity of particle A. From the problem, we know:
\[
v_A = 3v_p
\]
Let \( m_A \) be the mass of particle A. The ratio of the de Broglie wavelengths is:
\[
\frac{\lambda_A}{\lambda_p} = \frac{h/(m_A v_A)}{h/(m_p v_p)} = \frac{m_p v_p}{m_A v_A}
\]
Substitute \( v_A = 3v_p \) into the equation:
\[
\frac{m_p v_p}{m_A \cdot 3v_p} = \frac{3}{2}
\]
Simplifying:
\[
\frac{m_p}{3 m_A} = \frac{3}{2}
\]
\[
m_A = \frac{2}{9} m_p
\]
Thus, the mass of particle A is \( \frac{2}{9} m_p \).