The energy stored in the capacitor U = 21​CV2 = 21​C×(200)2 = 2C × 104 J This energy is used to heat up the block. Let Δθ be the rise in temperature, then heat energy Q = msΔθ = 0.1 x 250 x 0.4 = 10J Now, 2C × 104 = 10 ⇒ C = 2×10410​ = 5 × 10−4 = 500 µF