How to calculate the size of main memory if the cache is 4-way set associative memory, cache memory size is 256KB and number of tag bits is 8

I'm not able to calculate the number of blocks, number of sets, block offset, etc., which will eventually help me calculate the size of main memory.

None of that is required.

Your cache size is $256KB$, which means it can store sequence of $\log_2(2^8.2^{10}) = 18$ bits (assuming it is byte addressable). You will need $\log_2 4 = 2$ of these bits to access a particular block in the $4$-way set. Hence, set number and block offset together take $16$ bits and as mentioned in your question, tag takes $8$ bits; which makes the total address space consist of $24$ bits.

Therefore, main memory size is $2^{8+16}B = 2^{24}B = 16MB$.

Actually, the logical address space size is at most 16 MB.

The number of tag bits is an essential part of the hardware. If you had 8MB, only 7 tag bits would be needed, but you would have eight to make it possible to increase the memory size.

And your processor most likely has virtual memory, and therefore most likely a logical address space that is substantially larger than RAM. You need enough tags for your logical address space, not the much smaller RAM.