Correct : a

Explanation:
To find the total number of bits required to store all tag values in a direct-mapped cache, we first need to break down the structure of the physical address and determine how many tags (lines) the cache contains.
1. Analyze the Physical Address Structure:
The processor generates a 20-bit memory address. In a direct-mapped cache, this address is split into three fields:

[ Tag Bits | Index (Line) Bits | Block (Byte) Offset Bits ] = 20 bits total
• Block Offset Bits:
    The cache block size is 16 bytes = 2⁴ bytes.
    Since the memory is byte-addressable, the number of bits needed for the block offset = 4 bits.

• Index (Line) Bits:
    To find the number of lines (slots) in the cache:
    Number of Cache Lines = Cache Size / Block Size
    Number of Cache Lines = 16K bytes / 16 bytes = 1K lines = 2¹⁰ lines.
    Therefore, the number of bits needed to index these lines = 10 bits.

• Tag Bits:
    Tag Bits = Total Address Bits - (Index Bits + Block Offset Bits)
    Tag Bits = 20 - (10 + 4) = 20 - 14 = 6 bits per cache line.

2. Calculate Total Bits Required for All Tags:
Each of the cache lines requires its own tag store.
    Total Tag Bits = Number of Cache Lines × Tag Bits per Line
    Total Tag Bits = 2¹⁰ × 6 bits = 6 × 2¹⁰ bits.

3. Conclusion:
The total storage capacity needed to hold all tag structures is exactly 6×2¹⁰ bits, which perfectly matches option (a).