A biconvex lens is formed by using two thin planoconvex lenses, as...

The biconvex lens is a combination of two thin plano-convex lenses in contact.
For the first plano-convex lens: $\mu_1 = 1.5$, $R_1 = 15$ cm, $R_2 = \infty$.
Using lens maker's formula:
$\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
$= (1.5 - 1) \left( \frac{1}{15} - \frac{1}{\infty} \right)$
$= 0.5 \times \frac{1}{15} = \frac{1}{30}$ cm$^{-1}$.
For the second plano-convex lens:
$\mu_2 = 1.2$, $R_1 = \infty$, $R_2 = -12$ cm.
Using lens maker's formula:
$\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
$= (1.2 - 1) \left( \frac{1}{\infty} - \frac{1}{-12} \right)$
$= 0.2 \times \frac{1}{12} = \frac{1}{60}$ cm$^{-1}$.
The equivalent focal length $F$ of the combination is given by:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \frac{1}{60}$
$= \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20}$ cm$^{-1}$.
Thus, $F = 20$ cm.
Given object distance $u = -30$ cm. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$
$\Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$.
So, $v = 60$ cm.
Magnification $m = \frac{v}{u} = \frac{60}{-30} = -2$.