We can draw the table for better understanding.
$$\begin{array}{l|cc} \hline & \textbf{Small box} & \textbf{Large box} \\\hline \text{Selling price} & x & 2x \\ \text{Selling price per gram} & k & \left( \frac{88}{100}\right)k \\ \text{Weight} & \frac{x}{k} & \dfrac{2x}{\left(\frac{88}{100}\right)k} = \left(\frac{25}{11}\right)\frac{x}{k} \\\hline \end{array}$$
$\therefore$ The weight of chocolate in the large box exceeds that in the small box $= \left[\dfrac{\left(\frac{25}{11}\right)\frac{x}{k}-\frac{x}{k}}{\frac{x}{k}}\right] \times 100\%$
$\qquad = \left(\frac{25-11}{11}\right) \times 100 \%= \left(\frac{14}{11}\right) \times 100 \%=127.2727\%\cong127\%.$
$\textbf{Shortcut Method:}$
We can also do that in a more simple way.
$$\begin{array}{l|cc} \hline & \textbf{Small box} & \textbf{Large box} \\\hline \text{Selling price} & 1000 & 2000 \\ \text{Selling price per gram} & 100 & 88 \\ \text{Weight} & 10 & \frac{2000}{88} = \frac{250}{11} \\\hline \end{array}$$
$\therefore$ The weight of chocolate in the large box exceeds that in small box $= \left(\frac{\frac{250}{11}-10}{10}\right) \times 100\%$
$\qquad = \left(\frac{250-110}{110}\right) \times 100\%= \left(\frac{140}{110}\right) \times 100\%= \left(\frac{14}{11}\right) \times 100\%= 127.2727 \cong 127\%.$
Correct Answer $:\text{B}$