Correct Answer - `q//4 pi in_(0) a^(2)` along OE.
In Fig, suppose four particles each of charge q are placed at the corners A,B,C and D of regular pentagon
If we were to put the same charge q at E also, then by symmetry, net field at O would be zero. Hence combined field at O due to charges at A, B, C, D is equal and opposite to the field, at O due to charge at E. (which si `q//4 pi in_(0) a^(2)`, along EO).
Hence the combined field at O due to four charges at A,B,C,D `= q//4 pi in_(0) a^(2)` along OE.