Answer: d) 20 000 000 000 007

 $$\underbrace{3\times3}_{=\ 9}\times\underbrace{22...23}_{\begin{array}{l}\text{must be a}\\
\text{multiple of 9}\end{array}}$$
A number is a multiple of $9$ when its digits add up to a multiple of $9$
$2+2+3=7$
$2+2+2+3=9$ so $2223$ is a multiple of $9$
Adding more $2$s, it will always be odd
$\therefore$ next possible multiple of $9$ will be $27$ with one $3$ and twelve $2$s

Number of $2$s in $22...23$ equals the number of $0$s in $20...07$
$\therefore$ need three $0$s or twelve $0$s (or twenty-one, etc)

a) six $0$s
b) eight $0$s
c) ten $0$s
d) twelve $0$s