Minimum number of capacitors each of $8\mu F$ and 250V used to make a composite capacitor of $16\mu F$ and 1000V are

Number of capacitors in each row $n=\frac{V_{\text{required}}}{V_{\text{each}}}$

$n=\frac{1000}{250}=4$

$C_{\text{each\hspace{0.17em}\hspace{0.17em}row}}^1=\frac{C_0}{n}=\frac{8}{4}=2\mu F$

Number of rows $m=\frac{C_{\text{required}}}{C_{\text{each\hspace{0.17em}row}}}=\frac{16\mu F}{2\mu F}=8$

Total number of capacitors $N=mn=8\times 4=32$