If the ratio of lengths, radii and Young's moduli of steel and brass wires in the figure are a, b and c respectively, then the corresponding ratio of increase in their lengths is :

Tension in Brass wire ($$F_b$$): It supports only the bottom mass $$2M$$, hence, $$F_b = 2Mg$$
Tension in Steel wire ($$F_s$$): It supports both masses $$M$$ and $$2M$$, hence, $$F_s = Mg + 2Mg = 3Mg$$

$$\frac{F_s}{F_b} = \frac{3Mg}{2Mg} = \frac{3}{2}$$

$$\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$$

$$\frac{\Delta L_s}{\Delta L_b} = \left( \frac{F_s}{F_b} \right) \times \left( \frac{L_s}{L_b} \right) \times \left( \frac{r_b}{r_s} \right)^2 \times \left( \frac{Y_b}{Y_s} \right)$$

$$\frac{\Delta L_s}{\Delta L_b} = \left( \frac{3}{2} \right) \times (a) \times \left( \frac{1}{b} \right)^2 \times \left( \frac{1}{c} \right)$$

$$\frac{\Delta L_s}{\Delta L_b} = \frac{3a}{2b^2c}$$