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$Let\angle CAB=2a\\ \angle ACB=2c\\ \angle ABC=2b\\ We\quad Know\quad \angle a+2\angle c=\angle AEB(Exterior\quad angle\quad of\quad triangle)\\ Similarily\quad \angle c+2\angle a=\angle OFB\\ But\quad as\quad quadrilateral\quad FOEB\quad is\quad cyclic\quad Thetefore\angle OFB+\angle OEB=180(Property\quad of\quad cyclic\quad quad.)\\ Therefore\\ \angle a+2\angle c+2\angle a+\angle c=180\\ 3(\angle a+\angle c)=180\\ \angle a+\angle c=60\\ Now\quad in\quad Triangle\quad ABC\quad \angle A+\angle B+\angle C=180\\ so,\angle B=180-2(60)\\ \angle B=60\quad so,\quad \angle b=30\\ But\quad \angle c=\angle OFE(from\quad same\quad chord)\\ Therefore\quad \angle OFE=30$

The problem can be solved intuitively by considering ABC as an equilateral triangle(a general rule must apply to a special case). Thus, FBE=60 deg FOE=120 deg (FOEB is cyclic in given as well as in case of equilateral triangle) OFE=OEF since OF=OE= median/3 so, OFG=30 deg=(180-120)/2 deg