  For a prime number $p$, let $A_3(p)= | \{ m \in \mathbb{N}: \exists
m_1,m_2,m_3 \in \mathbb{N},
\frac{m}{p}=\frac{1}{m_1}+\frac{1}{m_2}+\frac{1}{m_3} \} |$. In 2019 Luca and
Pappalardi proved that $x (\log x)^3 \ll \sum_{p \le x} A_{3}(p) \ll x (\log
x)^5$. We improve the upper bound, showing $\sum_{p \le x} A_{3}(p) \ll x (\log
x)^3 (\log \log x)^2$.
