Answer
$2.4\ \mu F$
Work Step by Step
Given:
$C_1 = 6\ \mu F$
$C_2 = 4\ \mu F$
Potential difference $\Delta V = 200 V$
When two capacitance's $C_1$ and $C_2$ are connected in series their equivalent capacitance is
$\frac{1}{C_{equ}} = \frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{C_{equ}} = \frac{1}{6\ \mu F}+\frac{1}{4\ \mu F} $
$\frac{1}{C_{equ}} =0.4166\ \mu F$
$C_{equ} = 2.4\ \mu F$