The sum of coefficients of all the integral powers of  $x$ in the expansion of $(1+2\sqrt{x}{)}^{40}$ is

$(1+2\sqrt{x}{)}^{40}=1+{}^{40}{C}_1(2\sqrt{x}{)}^1+{}^{40}{C}_2(2\sqrt{x}{)}^2+{}^{40}{C}_3(2\sqrt{x}{)}^3+...+{}^{40}{C}_{40}(2\sqrt{x}{)}^{40}$
Hence, all the odd terms will have integral powers of $x$.
Substituting $x=1$, we get
$3^{40}=a_0+a_1+a_2+a_3...+a_{40}$ ...(i)
Now consider,
$(1-2\sqrt{x}{)}^{40}=1-{}^{40}{C}_1(2\sqrt{x}{)}^1+{}^{40}{C}_2(2\sqrt{x}{)}^2-{}^{40}{C}_3(2\sqrt{x}{)}^3...+{}^{40}{C}_{40}(2\sqrt{x}{)}^{40}$
Taking $x$ as $1$, we get
$1=a_0-a_1+a_2-a_3...+a_{40}$ ...(ii)
Adding i and, ii we get
$3^{40}+1=2[a_0+a_2+a_4+a_6...+a_{40}]$
$\frac{3^{40}+1}{2}=a_0+a_2+a_4+a_6...+a_{40}$
Hence the sum of the coefficients, of integral powers of $x$ will be
$\frac{3^{40}+1}{2}$
Hence, option D is correct