Initial symmetrical current, load short circuit

Base: 500 MVA, 20 kV. The terminal voltage is $V_t = 1\angle 0^\circ$ pu.

Pre-fault load current

The load is purely resistive, so the current is in phase with $V_t$:

$$I_L = \frac{P}{V_t} = \frac{400/500}{1} = 0.8\angle 0^\circ\ \text{pu}$$

Internal (sub-transient) voltage

$$\begin{aligned} E'' &= V_t + jX''_dI_L = 1 + j0.2(0.8) \\ &= 1 + j0.16 = 1.0127\angle 9.09^\circ\ \text{pu} \end{aligned}$$

Fault current

When all three load phases are shorted, the terminals are at zero voltage and the generator sees only $X''_d$:

$$\begin{aligned} I'' &= \frac{E''}{jX''_d} = \frac{1 + j0.16}{j0.2} = 0.8000 - j5.0000\ \text{pu} \\ |I''| &= \sqrt{0.8^2 + 5^2} = 5.0636\ \text{pu} \end{aligned}$$

In amperes: $I_{base} = \dfrac{500 \times 10^6}{\sqrt{3} \times 20 \times 10^3} = 14433.8$ A, so $|I''| = 5.0636 \times 14433.8 \approx 73.09$ kA.

Answer: initial symmetrical rms current $= 5.0636$ pu (about 73.1 kA).