Two containers, each with a capacity of 160 cc, are involved in a process where half the volume of one container is transferred to the other three times. Initially, the first container contains sugar syrup, and the second container contains water. After the third transfer, what is the ratio of sugar syrup to water in the second container?

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To solve this problem:

1. **Initial Setup**:  
   - Each container has a total capacity of 160 cc, with half (80 cc) being transferred during each step.  
   - Initially:  
     - Container 1 contains 160 cc of sugar syrup.  
     - Container 2 contains 160 cc of water.

2. **Step-by-Step Process**:  
   - **Step 1**: Transfer 80 cc from Container 1 (sugar syrup) to Container 2 (water).  
     - Container 1 now has **80 cc sugar syrup**.  
     - Container 2 now has **80 cc sugar syrup + 80 cc water** (total = 160 cc).

- **Step 2**: Transfer 80 cc from Container 2 (mixture) back to Container 1.  
     - The transferred liquid contains equal parts sugar syrup and water (40 cc each).  
     - After transfer:  
       - Container 1 now has **120 cc sugar syrup + 40 cc water**.  
       - Container 2 now has **40 cc sugar syrup + 120 cc water**.

- **Step 3**: Transfer another 80 cc from Container 1 (mixture) to Container 2.  
     - The transferred liquid contains **60 cc sugar syrup + 20 cc water** (proportional mixing).  
     - After transfer:  
       - Container 1 now has **60 cc sugar syrup + 20 cc water**.  
       - Container 2 now has **100 cc sugar syrup + 100 cc water**.

3. **Final Ratio in Container 2**:  
   The ratio of sugar syrup to water in Container 2 is:  
   $$
   	ext{Sugar Syrup : Water} = \frac{100}{120} = \frac{5}{6}.
   $$

Thus, the correct answer is **5:6**.

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