In the given circuit, the potential difference across $3\ \mu F$ capacitor will be:

both capacitor and battery are in series so

$V_{eq}=20-4=16v$

$C_eq=\dfrac{3 \times 5}{5+3}=\dfrac{15}{8}$

$q=C_{eq} \times V_{eq}=\dfrac{15}{8} \times16=30$

$V_{3}=\dfrac{q}{c}=\dfrac{30}{3}=10 \,v$