a
for first lens \(=\frac{1}{v_{1}}-\frac{1}{-x}=\frac{1}{1} \Rightarrow v_{1}=\frac{x}{x-1}\)

also magnification \(\left| m _{1}\right|=\left|\frac{ v _{1}}{ u _{1}}\right|=\frac{1}{ x -1}\)

for \(2^{nd}\) lens this is acting as object

so \(u _{2}=-\left(20- v _{1}\right)=-\left(20-\frac{ x }{ x -1}\right)\)

and \(v _{2}=-25 cm\)

angular magnification \(\left| m _{ A }\right|=\left|\frac{ D }{ u _{2}}\right|=\frac{25}{\left| u _{2}\right|}\)

Total magnification \(m = m _{1} m _{ A }=100\)

\(\left(\frac{1}{x-1}\right)\left(\frac{25}{20-\frac{x}{x-1}}\right)=100\)

\(\frac{25}{20(x-1)-x}=100 \Rightarrow 1=80(x-1)-4 x\)

\(\Rightarrow 76 x=81 \Rightarrow x=\frac{81}{76}\)

\(\Rightarrow u _{2}=-\left(20-\frac{81 / 76}{81 / 76-1}\right)=\frac{-19}{5}\)

now by lens formula

\(\frac{1}{-25}-\frac{1}{-19 / 5}=\frac{1}{f_{e}} \Rightarrow f_{e}=\frac{25 \times 19}{106} \approx 4.48 cm\)