Explanation

(c)
Volume of monobasic acid = $10{cm}^{3}$
Normality of monobasic acid = 0.1 N
Volume of NaOH solution= $15{cm}^{3}$
Normality of NaOH solution =?
$\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N}_{1}\ V_{1}\ \ = \ {\ \ N}_{2}\ V_{2}$
(for monobasic acid) (for NaOH)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ 10\ \times \ \ 0.1\ N\ = 15 \times N_{2}$
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N}_{2} = \frac{1N}{15} = 0.066\ N$