b$.$ In slide $10$ of lecture $23,$ we discussed how the barrier for nucleation could be reduced by having nucleation occur heterogeneously. For production of $Al$ castings, it is common practice to introduce "grain refiners" in the Al melt to increase the number of nuclei. These grain refiners are solid particles on which the nucleus makes an interface with very low energy $($i$.$e$.,{}_{Sl}$ in the figure in slide $10$ is very small$),$ such that the heterogeneous nucleation barrier $(G_{\text{h}\text{e}\text{t}})$ is much smaller than that for homogeneous nucleation $(G_{\text{h}\text{o}\text{m}}^{**}).$ The ratio of the heterogeneous nucleation rate to the homogeneous nucleation rate can be written as:

$\frac{N_{\text{h}\text{e}\text{t}}^{}}{N_{\text{h}\text{o}\text{m}}^{}}=K_1^{\text{h}\text{e}\text{t}}\frac{\text{e}\text{x}\text{p}(-\frac{G_{\text{h}\text{e}\text{t}}^{**}}{k}T)}{K_1^{\text{h}\text{o}\text{m}}}[exp(-\frac{G_{\text{h}\text{o}\text{m}}^{**}}{k}T)]$

where $K_1^{\text{h}\text{o}\text{m}}$ had $K_1^{\text{h}\text{e}\text{t}}$ are the number of sites available per unit volume for homogenous and heterogeneous nucleation, respectively. For homogenous nucleation, $K_1^{\text{h}\text{o}\text{m}}$ can be computed as the atomic density $($since every site in the bulk liquid is a potential site for homogeneous nucleation$),$ and takes a value of $K_1^{\text{h}\text{o}\text{m}}=53\mathrm{.}0n{m}^{-3}$ for Al$.$ Assume that the grain refiners reduce the nucleation barrier by a factor of $100,$ i$.$e$.,\frac{G_{\text{h}\text{e}\text{t}}^{**}}{}{G}_{\text{h}\text{o}\text{m}}^{**}=0\mathrm{.}01.$ Given this, and your answer for part $($a$),$ compute the density of grain refiners in units of $m^{-3})$ required for the heterogeneous nucleation rate to be $100$ times faster than the homogeneous nucleation rate in $Al,$ at $T=100K.$

Please assume the answer for part a$)$ as r$*($ in nm$)$ and theta G$($in eV$)$ and give a formula that can compute.