How should I measure out a solution given in nanorods per ml to make a solution of a set number of mols?
In order to avoid any possible misunderstanding, we should first clarify that “nanorods” (and thus also “nanorods per millilitre”) is not the name of a unit. A nanorod is a very small (so-called nanoscale) object.

The number of nanorods in one millilitre of the given solution is $N_\text{nanorods}=6.3\times10^{11}$, i.e.

$$\frac{N_\text{nanorods}}{V_\text{solution}}=\frac{6.3\times10^{11}}{1\ \mathrm{ml}}=6.3\times10^{14}\ \mathrm{l^{-1}}$$
where $V_\text{solution}$ is the volume of the solution.

Since

$$c=\frac nV$$

and

$$N_{\mathrm A}=\frac Nn=6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$

where $c$ is concentration, $n$ is amount of substance, and $N_{\mathrm A}$ is the Avogadro constant, the concentration of nanorods in the given solution is

$$\begin{align}c_\text{nanorods}&amp;=\frac{n_\text{nanorods}}{V_\text{solution}}\\[6pt]&amp;=\frac1{N_{\mathrm A}}\cdot\frac{N_\text{nanorods}}{V_\text{solution}}\\[6pt]&amp;=\frac1{6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}}\cdot6.3\times10^{14}\ \mathrm{l^{-1}}\\[6pt]&amp;=1.0\times10^{-9}\ \mathrm{mol\ l^{-1}}\\[6pt]&amp;=1.0\times10^{-6}\ \mathrm{mmol\ l^{-1}}\end{align}$$

The given nanorods are approximately cylinders with a length of $l=40\ \mathrm{nm}=4.0\times10^{-8}\ \mathrm m$ and a diameter of $d=10\ \mathrm{nm}=1.0\times10^{-8}\ \mathrm m$, i.e. a radius of $r=5\times10^{-9}\ \mathrm m$.

Therefore, the volume of one nanorod is approximately

$$\begin{align}V_\text{nanorod}&amp;=\pi\cdot r^2\cdot l\\[6pt]&amp;=\pi\times\left(5\times10^{-9}\ \mathrm m\right)^2\times4.0\times10^{-8}\ \mathrm m\\[6pt]&amp;=3.14\times10^{-24}\ \mathrm{m^3}\end{align}$$

The density of gold is $\rho=19.32\ \mathrm{g\ cm^{-3}}=1.932\times10^4\ \mathrm{kg\ m^{-3}}$. Thus, the mass of gold in one nanorod is approximately

$$\begin{align}m_{\ce{Au}}&amp;=\rho\cdot V_\text{nanorod}\\[6pt]&amp;=1.932\times10^4\ \mathrm{kg\ m^{-3}}\times3.14\times10^{-24}\ \mathrm{m^3}\\[6pt]&amp;=6.07\times10^{-20}\ \mathrm{kg}\end{align}$$

The molar mass of gold is $M_{\ce{Au}}=196.97\ \mathrm{g\ mol^{-1}}=0.196\,97\ \mathrm{kg\ mol^{-1}}$. Thus, the amount of gold in one nanorod is approximately

$$\begin{align}n_{\ce{Au}}&amp;=\frac{m_{\ce{Au}}}{M_{\ce{Au}}}\\[6pt]&amp;=\frac{6.07\times10^{-20}\ \mathrm{kg}}{0.196\,97\ \mathrm{kg\ mol^{-1}}}\\[6pt]&amp;=3.08\times10^{-19}\ \mathrm{mol}\end{align}$$

Note that this amount is equivalent to only about $190\,000$ gold atoms since

$$\begin{align}N_{\ce{Au}}&amp;=N_\mathrm A\cdot n_{\ce{Au}}\\[6pt]&amp;=6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}\times3.08\times10^{-19}\ \mathrm{mol}\\[6pt]&amp;=1.9\times10^5\end{align}$$

Therefore, the concentration of gold in the given solution can be estimated as follows:

$$\begin{align}c_{\ce{Au}}&amp;=n_{\ce{Au}}\cdot\frac{N_\text{nanorods}}{V_\text{solution}}\\[6pt]&amp;=3.08\times10^{-19}\ \mathrm{mol}\times6.3\times10^{14}\ \mathrm{l^{-1}}\\[6pt]&amp;=1.9\times10^{-4}\ \mathrm{mol\ l^{-1}}\\[6pt]&amp;=0.19\ \mathrm{mmol\ l^{-1}}\end{align}$$

or

$$\begin{align}c_{\ce{Au}}&amp;=N_{\ce{Au}}\cdot c_\text{nanorods}\\[6pt]&amp;=1.9\times10^5\times1.0\times10^{-9}\ \mathrm{mol\ l^{-1}}\\[6pt]&amp;=1.9\times10^{-4}\ \mathrm{mol\ l^{-1}}\\[6pt]&amp;=0.19\ \mathrm{mmol\ l^{-1}}\end{align}$$