The $$\mathrm{pK}_{\mathrm{a}_1}$$ and $$\mathrm{pK}_{\mathrm{a}_2}$$ of an amino acid are 2.3 and 9.7 respectively. The iso electric point of the amino acid is

For an amino acid with no ionizable side chain, the isoelectric point is the average of the two given values. $$pI=\frac{pK_{a_1}+pK_{a_2}}{2}$$ $$pI=\frac{2.3+9.7}{2}$$ $$pI=\frac{12.0}{2}=6.0$$ On checking the options, $6.0$ matches option $A$.