Answer : $512 MB$
The size of memory required at cache controller to store the metadata = $2KB = 2^{11}B$
Tag entry size in metadata = $x$ tag bits +$1$ modified bit +$1$ valid bit
#Cache Lines given = $1K = 2^{10}$
Block Size = $32 = 2^5B$
Given, Total size of metadata = $2^{11} * 8bits = 2^{14}b$
Total size of metadata = Tag entry size ×Number of cache blocks = $(x+1+1)* 2^{10}$
Hence, $2^{14} = (Tag+1+1)*2^{10} \implies Tag = 14 bits$
Thus Size of Main Memory address = Tag + Line Offset + Block Size = $14 + 10+5 = 29bits$
Therefore, the size of MM = $2^{29} = 512 MB$ (ANS)