Right answer is (b) 3.47 V
To explain I would say: \(X_C = \frac{1}{jCω} = \frac{-j}{2 × 3.14 × 100 × 10^3 × 10 × 10^{-12}}\)
Applying KCL at the node,
\(\frac{V_a-10}{100} + \frac{V_a}{100} + \frac{V_a}{500} + \frac{V_a}{-j159}\)
∴ Va = 4.37∠-15.95°.