Since, $a_r$ is the coefficient of $x^{10-r}$ in the binomial expansion of ${\left(1+x\right)}^{10}$, therefore

$a_r={}^{10}C_{10-r}={}^{10}C_r$

$a_{r-1}={}^{10}C_{r-1}$

So,

$\frac{a_r}{a_{r-1}}=\frac{{}^{10}C_r}{{}^{10}C_{r-1}}$

$\Rightarrow \frac{a_r}{a_{r-1}}=\frac{10!}{r!\times \left(10-r\right)!}\times \frac{\left(r-1\right)!\times \left(11-r\right)!}{10!}$

$\Rightarrow \frac{a_r}{a_{r-1}}=\frac{11-r}{r}$

Now,

$\sum _{r=1}^{10}{r}^3{\left(\frac{a_r}{a_{r-1}}\right)}^2=\sum _{r=1}^{10}{r}^3{\left(\frac{11-r}{r}\right)}^2$

$\Rightarrow \sum _{r=1}^{10}{r}^3{\left(\frac{a_r}{a_{r-1}}\right)}^2=\sum _{r=1}^{10}r{\left(11-r\right)}^2$

$\Rightarrow \sum _{r=1}^{10}{r}^3{\left(\frac{a_r}{a_{r-1}}\right)}^2=\sum _{r=1}^{10}r\left(121+r^2-22r\right)$

$\Rightarrow \sum _{r=1}^{10}{r}^3{\left(\frac{a_r}{a_{r-1}}\right)}^2=\sum _{r=1}^{10}\left(121r+r^3-22r^2\right)$

$\Rightarrow \sum _{r=1}^{10}{r}^3{\left(\frac{a_r}{a_{r-1}}\right)}^2=\frac{121\times 10\times 11}{2}+{\left(\frac{10\times 11}{2}\right)}^2-\frac{22\times 10\times 11\times 21}{6}$

$\Rightarrow \sum _{r=1}^{10}{r}^3{\left(\frac{a_r}{a_{r-1}}\right)}^2=6655+3025-8470$

$\Rightarrow \sum _{r=1}^{10}{r}^3{\left(\frac{a_r}{a_{r-1}}\right)}^2=1210$