Using the Binomial Theorem:
${{(x+y)^{n}=\sum_{r=0}^{n}\binom{n}{r}x^{n-r}y^{r}} }$
$(1+x)^{3}(2+x^{2})^{10} = \sum_{r=0}^{3}\binom{3}{r}(1)^{3-r}(x)^{r}\cdot \sum_{k=0}^{10}\binom{10}{k}(2)^{10-k}(x^{2})^{k}$
$(1+x)^{3}(2+x^{2})^{10} = \sum_{r=0}^{3}\binom{3}{r}(x)^{r}\cdot \sum_{k=0}^{10}\binom{10}{k}(2)^{10-k}(x^{2})^{k}$
Case$1:$If $r=3 , k=0$ we get $x^{3}$
So, Coefficient of $x^{3}$ is =$\binom{3}{3}$$\cdot \binom{10}{0}(2^{10})$ = $2^{10}$
Case$2:$If $r=1 , k=1 $ we get $x^{3}$
So, Coefficient of $x^{3}$ is = $\binom{3}{1}\cdot \binom{10}{1}(2)^{9}$ = $30\times 2^{9}$
So, finally Coefficient of $x^{3}$ is $ = 2^{10} + 30 \times 2^{9}$
$\qquad \qquad = 2^{9}( 2 + 30 )$
$\qquad \qquad = 2^{9}\times 32$
$\qquad \qquad=2^{9}\times 2^{5}$
$\qquad \qquad=2^{14}$
Correct Answer: A