In the circuit shown, initially there is no charge on capacitors...

For option $( C )$ , refer circuit diagram below

Just after closing of switch charge on all the capacitors $\left({C}_{1}, {C}_{3}and{C}_{4}\right) = 0$
$∴$ Replace all capacitors with wire, then circuit will get reduced as shown below.

$i = \dfrac{5}{70 + 100 + 30} = \dfrac{5}{200} = 25mA ⇒ ( C )$ is correct
Now ${S}_{1}$ is kept closed for long time circuit is in steady state

$⇒$ Apply KVL in loop, start from battery
$+ 5 - \dfrac{2}{80} - \dfrac{2}{10} - \dfrac{2}{80} = 0$
$⇒ \dfrac{10.2}{80} = 5$
$⇒ 2 = 40\muC$
$⇒$ Potential difference across ${C}_{1}$
$\Delta{V}_{{C}_{1}} = \dfrac{2}{10} = \dfrac{40}{10} = 4volt$
$⇒ ( D )$ is correct.
For option $( B ) ⇒$ Now just after closing of ${S}_{2}$ charge on each capacitor remain same,
Refer circuit below -

KVL in left loop, starting from battery of $10$ volts,
$10 - 30x - \dfrac{40}{10} - 70y = 0$
$⇒ 30x + 70y = 6$ ...(i)
Now, apply KVL in right loop starting from $80\muF$ capacitor
$⇒ - \dfrac{40}{80} + 5 + \left(x - y\right)30 - \dfrac{40}{80} + \left(x - y\right) \times 100 - 10 + x \times 30 = 0$
$⇒ 160x - 130y - 6 = 0$ ...(ii)
Now, solving equation (i) and (ii) we get,
$y = \dfrac{96}{1510}$
and $x = 0.05amp . ⇒ ( B )$ is incorrect.