A power station generates 1.0 MW at 10 kV…

Step 1. The allowed heat. $0.010 \times 1.0 \times 10^6\ \text{W} = 1.0 \times 10^4\ \text{W}$.

Step 2. The largest current. $I^2r = 1.0 \times 10^4$ W, so $I^2 = \dfrac{1.0 \times 10^4\ \text{W}}{16\ \Omega} = 625\ \text{A}^2$ and $I = 25$ A.

Step 3. The least voltage. We want $V$, so $V = \dfrac{P}{I} = \dfrac{1.0 \times 10^6\ \text{W}}{25\ \text{A}} = 4.0 \times 10^4\ \text{V} = 40$ kV.

Step 4. The turns ratio. The voltage goes from $10$ kV to $40$ kV, so $\dfrac{N_s}{N_p} = \dfrac{40\ \text{kV}}{10\ \text{kV}} = 4$. With $N_p = 500$ the secondary has $N_s = 4 \times 500 = 2000$ turns.