Step 1: Same heat flows through every rod.
The three rods AB, BC, CD are joined end to end (in series). In steady state the heat current $\dot{Q}$ is identical in all of them, and for each rod $\dot{Q} = \dfrac{K A \,\Delta T}{L}$, so $\Delta T = \dfrac{\dot{Q}}{A}\cdot\dfrac{L}{K}$.
Step 2: Get the lengths.
Each semicircular rod has length $\pi R = \dfrac{22}{7}\times 14 = 44\,\text{cm}$, so $L_{AB} = L_{CD} = 44\,\text{cm}$. The straight middle rod has $L_{BC} = 22\,\text{cm}$.
Step 3: Write the conductivities.
Given the ratio $1:2:3$, take $K_{AB} = k$, $K_{BC} = 2k$, $K_{CD} = 3k$. The area $A$ is the same for all.
Step 4: Use the middle rod to fix the constant.
Since $\Delta T \propto \dfrac{L}{K}$ (the common factor $\dot{Q}/A$ is shared), let $\Delta T = c\,\dfrac{L}{K}$. For BC, $30 = c\cdot\dfrac{22}{2k} = c\cdot\dfrac{11}{k}$, so $\dfrac{c}{k} = \dfrac{30}{11}$.
Step 5: Temperature drop across AB.
$\Delta T_{AB} = c\cdot\dfrac{44}{k} = \dfrac{c}{k}\times 44 = \dfrac{30}{11}\times 44 = 120^\circ\text{C}.$
Step 6: Temperature drop across CD.
$\Delta T_{CD} = c\cdot\dfrac{44}{3k} = \dfrac{c}{k}\times\dfrac{44}{3} = \dfrac{30}{11}\times\dfrac{44}{3} = 40^\circ\text{C}.$ So the two drops are $120^\circ\text{C}$ and $40^\circ\text{C}$.
\[ \boxed{120^\circ\text{C},\ 40^\circ\text{C}} \]