By how much per cent females reading newspaper E are more than the males reading newspaper C?

Number of males who read newspaper C

$\Rightarrow \frac{3}{8} \times\left(20000 \times \frac{16}{100}\right)=\frac{3}{8} \times 3200=1200$

Number of females who read newspaper $\mathrm{E}$

$\Rightarrow \frac{7}{11} \times\left(20000 \times \frac{22}{100}\right)=\frac{7}{11} \times 4400=2800$

$\therefore$ Required per cent $=\left(\frac{2800-1200}{1200}\right) \times 100$

$=\frac{1600}{12}=\frac{400}{3}=133.33 \%$