We prove that, for all primes $p>5$, positive integers $r$ and $m$ with $p\nmid m$ and $p\nmid (\genfrac{}{}{0ex}{}{2m{p}^{r-1}}{m{p}^{r-1}})$, there holds

$$\begin{array}{rl}& \frac{1}{m^4{p}^{4r}{(\genfrac{}{}{0ex}{}{2m{p}^{r-1}}{m{p}^{r-1}})}^3}(\sum _{k=0}^{m{p}^r-1}(21k+8){(\genfrac{}{}{0ex}{}{2k}{k})}^3-p\sum _{k=0}^{m{p}^{r-1}-1}(21k+8){(\genfrac{}{}{0ex}{}{2k}{k})}^3)\\ & \equiv -6\frac{H_{p-1}}{p^2}(\mathrm{m}\mathrm{o}\mathrm{d}{p}^2),\end{array}$$

where $H_n=\sum _{j=1}^n\frac{1}{j}$ ( $n=1,2,\cdots$) are the usual harmonic numbers. This partially confirms a conjecture by Z.-W. Sun.