The notation is important here.
If ${\bf F}$ is a continuous vector field defined over a surface oriented $S$ with a unit normal vector ${\bf n}$, then the surface integral of ${\bf F}$ on $S$ is given by $$\iint_{S}{\bf F}\cdot {\rm d}{\bf S}=\iint_{S}{\bf F}\cdot {\bf n}{\rm d}S=\iint_{S}{\bf F}\cdot \frac{{\bf r}_{u}\times {\bf r}_{v}}{|{\bf r}_{u}\times {\bf r}_{v}|}{\rm d}S=\iint_{R}\left[{\bf F}({\bf r}(u,v))\cdot \frac{{\bf r}_{u}\times {\bf r}_{v}}{|{\bf r}_{u}\times {\bf r}_{v}|} \right] |{\bf r}_{u}\times {\bf r}_{v}|{\rm d}A.$$
where $S$ is defined by a vectorial function ${\bf r}(u,v)$ and $R$ is the domain of the parameter. Hence,
$$\bbox[5px,#ffd]{\iint_{S}{\bf F}\cdot {\rm d}{\bf S}=\iint_{R}{\bf F}\cdot ({\bf r}_{u}\times {\bf r}_{v}){\rm d}A}$$
Notice that $S$ is the portion of the cone $z=\sqrt{x^{2}+y^{2}}$ and is inside a cylinder $x^2+y^2=1$. Then in cylindrical coordinates the cone is $z=r$, so using parameters $r,\theta$ we have the parametrization $${\bf r}(r,\theta)=(r\cos \theta,r\sin \theta,r),\quad 0\leqslant r\leqslant 1,\quad 0\leqslant \theta \leqslant 2\pi$$
Then $${\bf r}_{r}\times {\bf r}_{\theta}=(-r\cos \theta,-r\sin \theta,r).$$
Since the $z$-coordinate is positive, then ${\bf r}_{r}\times {\bf r}_{\theta}$ which has upward orientation.
Then $${\bf F}\cdot ({\bf r}_{r}\times {\bf r}_{\theta})={\bf F}({\bf r}(r,\theta))\cdot ({\bf r}_{r}\times {\bf r}_{\theta})=(2,5,3)\cdot (-r\cos \theta,-r\sin \theta,r)=-2r\cos \theta-5r\sin \theta+3r.$$
Hence,
$$\iint_{S}{\bf F}\cdot d{\bf S}=\int_{0}^{2\pi}\int_{0}^{1}(-2r\cos \theta-5r\sin \theta+3r){\rm d}r{\rm d}\theta=3\pi$$
Therefore the answer is $3\pi$, this represents the flow of ${\bf F}$ through $S$.
