When connected in series, the sum of the reciprocals of each capacitance is equal to the reciprocal of the equivalent capacitance.

(1/C_{1}) + (1/C_{2}) = (1/C_{eq})

If we subtract (1/C_{1}) from both sides of the equation:

(1/C_{2}) = (1/C_{eq}) - (1/C_{1})

(1/C_{2}) = (1/(1.50*10^{-5})) - (1/(5.00*10^{-5}))

Then we multiply the first fraction by 10/10 and the second fraction by 3/3 to obtain a common denominator so we can add the fractions. (We can always multiply by 1) :

(1/C_{2}) = (10/(1.5*10^{-4})) - (3/(1.5*10^{-4}))

(1/C_{2}) = (7/(1.5*10^{-4}))

And take the reciprocal of the result:

C_{2} = ((1.5*10^-4)/7) = 2.14 * 10^{-5} F