$(a)\;29.90 mL\qquad(b)\;29 mL\qquad(c)\;29.56 mL\qquad(d)\;30.34 mL$

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Since, The molar equivalent of a solution does not change on dilution,

Molar equivalent of conc. $HNO_3$ = Molar equivalent of dilute $HNO_3$

We have, Normality of $HNO_3$ = 15.78

$\therefore 2\times15.78 = V\times1$

$\therefore$V = 31.56 mL

$\therefore$ Volume of water added = 31.56 - 2 = 29.56 mL

Hence answer is (c)

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