When does a star rise with the sun?
The formula you gave is to find the hour angle of the star while setting, not the setting time itself. Suppose the hour angle of star is $HA\star$, and $RA=\alpha$,then the Local Sidereel Time is given by $LST=HA\star+\alpha$. At the time of setting, $HA\star=10^h1^m30.2^s$. Thus, $LST=13^h16^m30.2^s$at the time of setting.

Now, $RA\odot=6^h$ on $\text{June} 21^{st}$ and $HA_\odot =\arccos(\tan(23.43^{\circ}))=4^h17^m17^s$ at sunset. Thus the star sets with the sun after $\text{ June} 21^{st}$. Since $RA_\odot$ increases approximately by $2^h$ every month, we can assume that the star sets with the sen near $\text{July }21^{st}$. Let it set after $n$ days. At $\text{July }21^{st}$, $\delta_\odot = 20.09^{\circ}$ which gives us $HA_\odot=4^h34^m11^s$, and $RA_{\odot}=8^h9^m44^s$. The $HA_\odot$ at sunset is approximately the same for nearby dates. Since $\delta_\odot$ is also near the maximum, we can approximate $\Delta RA_\odot= \omega_\odot n$. Thus,

$$13^h16^m30.2^s=8^h9^m44^s + \omega_\odot n +  4^h34^m11^s$$
$$ \omega_\odot n =0.5433^h$$
$$n=8$$

At  $\text{July } 29^{th}$, $\delta_\odot=18.23^{\circ}$, $HA_\odot= 4^h43^m6^s$, $RA_\odot=8^h42^m12^s$ which gives us $LST=13^h25^m18^s$, which is a bit more than expected. Trial and error for few days before gives us that on $\text{July } 26^{th}$, the sun sets at LST $13^h15^m9^s$ which would be the most appropriate answer.

Thus, the sun sets along with the star at $\text{July } 26^{th}$