In which two states is the number of IT graduates, as a percentage of the total number of graduates in that state, most nearly the same?

Percentage of IT graduates in state A,

= $$\frac{31}{46 + 31 + 23 + 22} \times 100 = \frac{3100}{122} = 25.4$$%

Percentage of IT graduates in state B,

= $$\frac{38}{41 + 38 + 28 + 28} \times 100 = \frac{38}{135} = 28.14$$%

Percentage of IT graduates in state C,

= $$\frac{22}{62 + 34 + 26 + 23} \times 100 = \frac{22}{145} = 15.17$$%

Percentage of IT graduates in state D,

= $$\frac{18}{59 + 45 + 31 + 18} \times 100 = \frac{18}{153} = 11.76$$%

Percentage of IT graduates in state E,

= $$\frac{27}{38 + 28 + 24 + 12} \times 100 = \frac{27}{102} = 26.47$$%

Percentage of IT graduates in state F,

= $$\frac{16}{54 + 25 + 16 + 11} \times 100 = \frac{16}{106} = 15.09$$%

$$\therefore$$ Percentage of IT graduates in C and F is mostly same.

Hence, option A is the correct answer.