Apparent shift in position due to a glass slab is given by $d=t\left(1-\frac{1}{\mu }\right)$.

$\Rightarrow d=1.4\left(1-\frac{5}{7}\right)=0.4 \mathrm{cm}$

For refraction without glass slab, $u=-20 \mathrm{cm}, v=5 \mathrm{cm}$. Applying lens equation,

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{5}+\frac{1}{20} \\ \Rightarrow f=4 \mathrm{cm} \end{gathered}$$

For refraction with glass slab inserted, $v=5-0.4=4.6 \mathrm{cm}, f=4 \mathrm{cm}$. Applying lens equation,

$$\begin{gathered} \frac{1}{4}=\frac{1}{4.6}-\frac{1}{u} \\ \Rightarrow u=30.7 \mathrm{cm} \end{gathered}$$