Given, $m_1=4$ and $m_2=3$
Also, $\quad u_2-u_1=2 \mathrm{~cm}$
We know that, magnification produced by a lens is given as
$m=\frac{v}{u}$
$\Rightarrow \quad m_1=\frac{v_1}{u_1} \Rightarrow \frac{u_1}{v_1}=\frac{1}{m_1}=\frac{1}{4}$ $\ldots(\mathrm{i})$
and $\quad m_2=\frac{v_2}{u_2} \Rightarrow \frac{u_2}{v_2}=\frac{1}{m_2}=\frac{1}{3}$ $\ldots(\mathrm{ii})$
Using lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{u_1}{f}=\frac{u_1}{v_1}-1$
$\Rightarrow \quad \frac{u_1}{f}=\frac{1}{4}-1$...(iii) [from Eq. (i)]
Similarly, $\frac{u_2}{f}=\frac{u_2}{v_2}-1$
$\Rightarrow \quad \frac{u_2}{f}=\frac{1}{3}-1$...(iv) [from Eq. (ii)
Using Eqs. (iii) and (iv), we have
$\frac{u_2}{f}-\frac{u_1}{f}=\left(\frac{1}{3}-1\right)-\left(\frac{1}{4}-1\right)$
$\Rightarrow \quad \frac{1}{f}\left(u_2-u_1\right)=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$
$\Rightarrow f=\left(u_2-u_1\right) \times 12 \Rightarrow f=2 \times 12=24 \mathrm{~cm}$