Size of the Disk $=4\;\text{TB}=4\times 2^{40} = 2^{42} \ Bytes$
Size of a Block $=4\;\text{KB}=4\times 2^{10} = 2^{12} \ Bytes$
Number of Blocks present in the Disk $= Size \ of \ Disk / Size \ of \ Block = 2^{42} / 2^{12} = 2^{30}$
Number of Bits needed to represent address of one Block $= \lg 2^{30} \ bits = 30 \ bits$ $\approx$ $4 \ Byte$
Number of block addresses that can be stored in one block $= 2^{12} / 4 = 2^{10} = 1024$