time it takes for the voltage to achieve 7 V
In your circuit:



Since the current in a series circuit is everywhere the same at any instant, when S1 closes, the accumulation of charge in C1, in time, will be equal to the accumulation of charge in C2, but since their capacitances are different the voltages across them will be different. .

Then, since \$ Q=CV\$, and E3 is given as 7 volts,

for C2, $$ Q = C2 \times E3 =12\mu F \times 7V =84\mu C $$

and for C1, $$ V=\frac {Q}{C} = \frac{84\mu C}{56\mu F} = 1.5\text{ volts.}$$

The voltage across C2 is given as 7 volts, so E2 will be equal to 7 volts plus the voltage across C1, 1.5 volts, so $$ E2 = E3+1.5V = 8.5V $$

C1 and C2 are in series, so their equivalent capacitance is:

$$C_T = \frac{C1 \times C2}{C1+C2} \approx 9.9 \text{ microfarads.}$$

Our circuit now looks like this:


and we can get the time taken to charge Ct to 8.5 volts (which is the time taken to charge C2 to 7 volts) by solving:

$$\tau = k \ RC,  $$

where

$$ k = ln\frac{V}{V-V_T} = ln\frac{13V}{13V-8.5V} =1.06$$ 

which works out to:

$$\tau = 1.06\times\ 1.2\cdot10^{4}\Omega\times 9.9\cdot10^{-6}F \approx 126\text{ milliseconds}. $$

The proof:



and the LTspice circuit list:

Version 4
SHEET 1 880 680
WIRE 144 48 128 48
WIRE 224 48 144 48
WIRE 400 48 304 48
WIRE 480 48 400 48
WIRE 400 80 400 48
WIRE 400 176 400 144
WIRE 480 176 400 176
WIRE 128 208 128 48
WIRE 400 208 400 176
WIRE 128 304 128 288
WIRE 400 304 400 272
WIRE 400 304 128 304
WIRE 128 352 128 304
FLAG 128 352 0
FLAG 480 176 E3
FLAG 480 48 E2
FLAG 144 48 E1
SYMBOL voltage 128 192 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value PULSE(0 13 0 100n)
SYMBOL cap 384 80 R0
SYMATTR InstName C1
SYMATTR Value 56µ
SYMBOL cap 384 208 R0
SYMATTR InstName C2
SYMATTR Value 12µ
SYMBOL res 320 32 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 12k
TEXT 136 328 Left 2 !.tran 150m startup uic