The general term in the binomial expansion of ${\left(a+b\right)}^n$ is ${}^{n}C_R{a}^{n-R}{b}^R.$

Thus, the general term in the expansion of ${\left(2x^r+\frac{1}{x^2}\right)}^{10}$ is $={}^{10}C_R{\left(2x^r\right)}^{10-R}{\left(\frac{1}{x^2}\right)}^R$

$={}^{10}C_R\left(2^{10-R}{x}^{r\left(10-R\right)}\right)x^{-2R}$

Given, the constant term in the expansion is $180$

$\Rightarrow {}^{10}C_R·2^{10-R}=180 \dots \left(1\right)$ 

And $(10-R)r-2R=0$

$\Rightarrow r=\frac{2R}{10-R}$

$\Rightarrow r=\frac{2(R-10)}{10-R}+\frac{20}{10-R}$

$\Rightarrow r=-2+\frac{20}{10-R} \dots \left(2\right)$

Since $R$ is positive integer less than or equal to $10$ and $r$ is also an integer, hence, the value of $R$ is such that, $10-R$ must divide $20.$

This is possible, when the value of $10-R$ can be one from the divisors of $20$ i.e. $10-R$ can be from $1, 2, 4$ or $5$ and consequently, $R$ can be $9, 8, 6$ or $5$ respectively.

But for $R=9, 6 \& 5$ equation $\left(1\right)$ is  not satisfied, hence $R=8$

as ${}^{10}C_R{2}^{10-R}=180$

$\Rightarrow r=\frac{2\times 8}{10-8}=8.$