$1 \, L$ aqueous solution of $H_{2}SO_{4}$ contains $0.02 \, mmol$ of $H_{2}SO_{4}$. $50 \%$ of this solution is taken and diluted with deionized water to give $1 \, L$ solution $(A)$. In solution $(A)$, $0.01 \, mmol$ of $H_{2}SO_{4}$ is added. The total amount of $H_{2}SO_{4}$ in the final solution is $x \times 10^{-3} \, mmol$. The value of $x$ is:

• A
  $3$
• B
  $1$
• C
  $20$
• D
  $2$