1. Set the geometry

   Let the concave mirror be at the left and the convex lens be at the right, separated by $50\,\text{cm}$.

   • Focal length of concave mirror: $f_m = 10\,\text{cm}$
   • Focal length of convex lens in air: $f_L = 10\,\text{cm}$
   • Object is $15\,\text{cm}$ in front of the mirror.

   Since the final image is stated to be erect, we first trace the image formation in air and then in the medium.

2. Image formed by the concave mirror

   Using mirror formula: $\frac{1}{f_m} = \frac{1}{v_m} + \frac{1}{u_m}$

   For a concave mirror with object in front: $f_m = 10\,\text{cm}, \quad u_m = 15\,\text{cm}$

   So, $\frac{1}{10} = \frac{1}{15} + \frac{1}{v_m}$ $\frac{1}{v_m} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30}$ $v_m = 30\,\text{cm}$

   Thus the mirror forms a real image $30\,\text{cm}$ in front of it, i.e. between mirror and lens.

   Since mirror-lens separation is $50\,\text{cm}$, this image is at a distance $50 - 30 = 20\,\text{cm}$ to the left of the lens.

   So for the lens, the object distance is $u_L = 20\,\text{cm}$

3. Case 1: Setup in air

   Lens focal length in air is $10\,\text{cm}$.

   Using lens formula: $\frac{1}{f_L} = \frac{1}{v_L} - \frac{1}{u_L}$

   With Cartesian convention for a real object on left: $u_L = -20\,\text{cm}$, $f_L = +10\,\text{cm}$.

   Hence, $\frac{1}{10} = \frac{1}{v_L} - \left(-\frac{1}{20}\right)$ $\frac{1}{10} = \frac{1}{v_L} + \frac{1}{20}$ $\frac{1}{v_L} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}$ $v_L = 20\,\text{cm}$

   So the final image is real and to the right of the lens.

   Magnifications:

   • Mirror magnification: $m_m = -\frac{v_m}{u_m} = -\frac{30}{15} = -2$
   • Lens magnification: $m_{L1} = \frac{v_L}{u_L} = \frac{20}{-20} = -1$

   Therefore total magnification in air: $M_1 = m_m \cdot m_{L1} = (-2)(-1)=2$

   This is positive, so image is erect, consistent with the question.

4. Case 2: Setup in medium of refractive index $\dfrac{7}{6}$

   The focal length of the mirror does not change with medium: $f_m = 10\,\text{cm}$

   So mirror image remains the same: $v_m = 30\,\text{cm}$ and for the lens, object distance is still $u_L = 20\,\text{cm} \quad (\text{or } -20\,\text{cm in sign convention})$

5. New focal length of the lens in the medium

   Lens maker relation in a medium: $\frac{1}{f} = \left(\frac{n_{\ell}}{n_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

   In air: $\frac{1}{f_{air}} = (1.5-1)K = 0.5K$ where $K = \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

   Given $f_{air}=10\,\text{cm}$, $\frac{1}{10}=0.5K \Rightarrow K=\frac{1}{5}$

   In medium $n_m=\frac{7}{6}$, $\frac{n_{\ell}}{n_m} = \frac{1.5}{7/6} = \frac{3/2}{7/6} = \frac{9}{7}$

   Hence, $\frac{1}{f_{med}} = \left(\frac{9}{7}-1\right)K = \frac{2}{7}\cdot \frac{1}{5} = \frac{2}{35}$ $f_{med} = \frac{35}{2}=17.5\,\text{cm}$

6. Image formed by the lens in the medium

   Now use lens formula with $f=17.5\,\text{cm}, \quad u=-20\,\text{cm}$

   $\frac{1}{17.5} = \frac{1}{v} - \left(-\frac{1}{20}\right)$ $\frac{1}{17.5} = \frac{1}{v} + \frac{1}{20}$

   Therefore, $\frac{1}{v} = \frac{1}{17.5} - \frac{1}{20}$ $\frac{1}{v} = \frac{2}{35} - \frac{1}{20} = \frac{8-7}{140} = \frac{1}{140}$ $v = 140\,\text{cm}$

   Lens magnification in medium: $m_{L2} = \frac{v}{u} = \frac{140}{-20} = -7$

   Total magnification: $M_2 = m_m \cdot m_{L2} = (-2)(-7)=14$

7. Required ratio

   $\left|\frac{M_2}{M_1}\right| = \left|\frac{14}{2}\right| = 7$

8. Final answer

   $\boxed{7}$

The derived answer matches the stored correct answer.