Explanation

(b) Suppose $C = 8\ \mu F,\ C^{''} = 16\mu F$
And $V = 250V,\ V^{''} = 1000V$
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Suppose $m$ rows of given capacitors are connected in parallel and each row contains $n$ capacitors then potential difference across each capacitor $V = \frac{V^{''}}{n}$ and equivalent capacitance of network $C^{''} = \frac{mC}{n}$ on putting the values we get $n = 4$ and $m = 8$
$\therefore$ Total capacitors = $n \times m = 4 \times 8 = 32$
Short Trick : For such type of problems number of capacitors
$= \frac{C^{''}}{C} \times \left( \frac{V^{''}}{V} \right)^{2} = \frac{15}{8}\left( \frac{1000}{250} \right)^{2} = 32$