We have, 

$\left(1+ax+b{x}^2\right){\left(1-2x\right)}^{18}$

$\left(1+ax+b{x}^2\right)\left(1-{}^{18}C_{1}2x+{}^{18}C_2\left(2x^2\right)-{}^{18}C_3{\left(2x\right)}^3+{}^{18}C_4{\left(2x\right)}^4\dots \right)$

Coefficient of $x^3$ is $-{}^{18}C_3{\left(2\right)}^3+a·{}^{18}C_2{\left(2\right)}^2-b^{18}{C}_1\left(2\right)$

and coefficient of $x^4$ is $-{}^{18}C_4{\left(2\right)}^4-{}^{18}C_3{\left(2\right)}^{3}a+{}^{18}C_2{\left(2\right)}^{2}b$

Coefficient of $x^3$ and $x^4$ are zero.

$\therefore -{}^{18}C_3{\left(2\right)}^3+{}^{18}C_2{\left(2\right)}^2\left(a\right)-{}^{18}C_1\left(2\right)b=0$

$\Rightarrow \frac{4\times 17\times 16}{3\times 2}-17a+b=0$    $...\left(\mathrm{i}\right)$

and $80-\frac{32}{3}a+b=0$   $...\left(\mathrm{ii}\right)$

Solving Eqs. $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get

$a=16,b=\frac{272}{3}$