Correct Answer: $\frac{6}{49}$

Solution : Given: $\frac{\left(2 \frac{1}{3}+2 \frac{1}{2}-\frac{1}{6}\right) \div 2 \frac{1}{3} \times 5 \frac{2}{3} \div 1 \frac{2}{3} \text { of } 4 \frac{1}{4}}{3 \frac{1}{5} \div 4 \frac{1}{2} \text { of } 5 \frac{1}{3}+5 \frac{1}{3} \times \frac{3}{4} \div 2 \frac{2}{3}}$
= $\frac{(\frac{7}{3}+\frac{5}{2}-\frac{1}{6}) \div  \frac{7}{3} \times  \frac{17}{3} \div  \frac{5}{3} \text { of } \frac{17}{4}}{\frac{16}{5} \div \frac{9}{2} \times\frac{16}{3}+ \frac{16}{3} \times \frac{3}{4} \div \frac{8}{3}}$
= $\frac{(\frac{14+15-1}{6}) \times  \frac{3}{7} \times  \frac{17}{3} \div  \frac{85}{12}}{\frac{16}{5} \div \frac{24}{1}+ \frac{16}{3} \times \frac{3}{4} \times \frac{3}{8}}$
= $\frac{\frac{28}{6} \times  \frac{3}{7} \times  \frac{17}{3} \times  \frac{12}{85}}{\frac{16}{5} \times \frac{1}{24}+ \frac{16}{3} \times \frac{3}{4} \times \frac{3}{8}}$
= $\frac{\frac{16}{15}}{\frac{2}{15} +\frac{3}{2}}$
= $\frac{48}{49}$
Student's answer was $\frac{6}{7}$.
Difference = $\frac{48}{49}-\frac{6}{7}$ = $\frac{6}{49}$
Hence, the correct answer is $\frac{6}{49}$.