We are given that $\left(a+bx+cx^2\right)^{10} = \sum_{i=0}^{20} p_i x^i$, and we are given that $p_1 = 20$ and $p_2 = 210$.

We need to find the value of $2(a+b+c)$.

Using the multinomial theorem, we can express the expansion of $(a + bx + cx^2)^{10}$ as follows:

$$ \sum\limits_{k_1+k_2+k_3=10} {{10!} \over {{k_1}!{k_2}!{k_3}!}} a^{k_1} (bx)^{k_2} (cx^2)^{k_3} $$

Now we need to find the coefficients of $x^1$ and $x^2$ in the expansion:

For $x^1$ term, we have:

$$ k_2 = 1, k_1 = 9, k_3 = 0 $$

So,

$$ p_1 = {{10!} \over {9!1!0!}} a^9 b^1 = 10a^9 b $$

For $x^2$ term, there are two possibilities:

$$ k_2 = 2, k_1 = 8, k_3 = 0 \quad \text{and} \quad k_2 = 0, k_1 = 9, k_3 = 1 $$

So,

$$ p_2 = {{10!} \over {8!2!0!}} a^8 b^2 + {{10!} \over {9!0!1!}} a^9 c = 45a^8 b^2 + 10a^9 c $$

Now we are given $p_1 = 20$ and $p_2 = 210$. So, $$ 10a^9 b = 20 \implies a^9 b = 2 $$

and $$ 45a^8 b^2 + 10a^9 c = 210 $$

Now, divide the second equation by $a^8$: $$ 45b^2 + 10ac = 210 $$

We know that $a^9 b = 2$. Taking the $9^{th}$ root of both sides: $$ ab = \sqrt[9]{2} $$

Now, let $k = ab = \sqrt[9]{2}$. We can rewrite the equation for $x^2$ term as: $$ 45k^2 + 10k^9 = 210 $$

From the equation $ab = k = \sqrt[9]{2}$, we know that $a$ and $b$ are positive integers. Thus, $k = 2$ (as both $a$ and $b$ must be factors of 2). Now we have:

$$ a+b = 2 $$

and from the equation $a^9 b = 2$, we get $a = 1, b = 2$ or vice versa.

Now we need to find the value of $c$. We can use the equation for the $x^2$ term again:

$$ 45a^8 b^2 + 10a^9 c = 210 $$

Using $a=1$ and $b=2$, we get:

$$ 45(1)^8 (2)^2 + 10(1)^9 c = 210 \implies 180 + 10c = 210 \implies c = 3 $$

So, $a=1$, $b=2$, and $c=3$. Now, we need to find the value of $2(a+b+c)$:

$$ 2(a+b+c) = 2(1+2+3) = 2(6) = 12 $$

Therefore, the answer is $\boxed{12}$.