Answer
4
Work Step by Step
For series connection with n equivalent capacitors,
$\frac{1}{C_{eq}}=(\frac{1}{C})n=(\frac{1}{2.00\,\mu F})n$
$\implies C_{eq}=\frac{2.00\,\mu F}{n}$
Now, Energy $U=\frac{1}{2}CV^{2}$
$=\frac{1}{2}\times(\frac{2.00\,\mu F}{n})\times (10\,V)^{2}=25\,\mu J$
$\implies \frac{1}{n}\times1.00\,\mu F\times100\,V^{2}=25\,\mu J$
$n=\frac{1.00\,\mu F\times100\,V^{2}}{25\,\mu J}=4$
