If the page size in a 32-bit machine is 4K bytes then the size of page table is

In a 32-bit machine, the virtual address space is $2^{32}$ bytes. With a page size of 4K ($2^{12}$) bytes, the number of pages is $2^{32} / 2^{12} = 2^{20}$ pages. Assuming a standard page table entry size of 4 bytes, the total page table size is $2^{20} \times 4$ bytes = 4 MB.