The energy stored by a system of capacitors is given by
[tex]U= \frac{1}{2}C_{eq} V^2 [/tex]
where Ceq is the equivalent capacitance of the system, and V is the voltage applied.
In the formula, we can see there is a direct proportionality between U and C. This means that if we want to increase the energy stored by 4 times, we have to increase C by 4 times, if we keep the same voltage.
Calling [tex]C_1 = 270 pF[/tex] the capacitance of the original capacitor, we can solve the problem by asking that, adding a new capacitor with [tex]C_x[/tex], the new equivalent capacitance of the system [tex]C_{eq}[/tex] must be equal to [tex]4C_1[/tex]. If we add the new capacitance X in parallel, the equivalent capacitance of the new system is the sum of the two capacitance
[tex]C_{eq} = C_1 + C_x[/tex]
and since Ceq must be equal to 4 C1, we can write
[tex]C_1+C_x = 4C_1[/tex]
from which we find
[tex]C_x=3C_1=3 \cdot 270 pF=810 pF[/tex]
