Each vector has its magnitude & direction then we have
$$\bar r=|r|\hat r=r\hat r$$
Since, the unit radial vector $\hat r$ & the unit normal vector $\hat n$ to the spherical surface are in the same direction i.e. the angle between these vectors $\alpha=0^{o} \space$ hence we have
$$\hat r \cdot \hat n=|\hat r| \cdot |\hat n|cos \alpha=1 \cdot 1 cos0^{o}=1$$
And an elementary area $dS$ on the spherical surface with a radius $r\space$ is given as
$$dS=(rd \theta)\cdot(rsin\theta d\phi)=r^2sin\theta d\theta d\phi$$
Where, $\theta$ is angle of longitude & $\phi$ is angle of latitude
Now, substituting the corresponding values in left hand side of the expression we get the solid angle
$$=\iint_S\frac{\bar r\cdot \hat n \space dS}{r^3}=\iint_S\frac{r(\hat r\cdot \hat n) \space r^2sin\theta d\theta d\phi}{r^3}=\iint_S\frac{r^3(1)\space sin\theta d\theta d\phi}{r^3}=\iint_Ssin\theta d\theta d\phi$$
It is obvious that the solid angle is a dimensionless quantity having SI unit Ste-radian (sr).