Correct Answer: $7$

Solution : Given: $\frac{28}{21} \times \frac{96}{4} \div \frac{125}{5} \times \frac{42}{192}+\frac{297}{125} \times \frac{875}{3} \div \frac{99}{4} \times \frac{6}{25}$
= $\frac{\frac{4}{3} × \frac{96}{4}}{\frac{125}{5}}$ × $\frac{42}{192}$ + $\frac{\frac{297}{125}×\frac{875}{3}}{\frac{99}{4}}$ × $\frac{6}{25}$
= $\frac{\frac{4}{3} × 24}{\frac{125}{5}}$ × $\frac{42}{192}$ + $\frac{\frac{297}{125}×\frac{875}{3}}{\frac{99}{4}}$ × $\frac{6}{25}$
= $\frac{ 32}{25}$ × $\frac{7}{32}$ + $\frac{\frac{297}{125}×\frac{875}{3}}{\frac{99}{4}}$ × $\frac{6}{25}$
= $\frac{ 32}{25}$ × $\frac{7}{32}$ + $\frac{{693}}{\frac{99}{4}}$ × $\frac{6}{25}$
= $\frac{ 7}{25}$ + $\frac{{693}}{\frac{99}{4}}$ × $\frac{6}{25}$
= $\frac{ 7}{25}$ + 28 × $\frac{6}{25}$
= $\frac{7}{25}$ + $\frac{168}{25}$
= $\frac{175}{25}$ = 7
Hence, the correct answer is $7$.