1. Find the resistance of the heater

Given rated power and voltage of heater: $P = 1000\,\text{W}, \qquad V = 100\,\text{V}$

Using $P = \frac{V^2}{R_h}$ we get heater resistance $R_h = \frac{V^2}{P} = \frac{100^2}{1000} = 10\,\Omega.$

So the heater has resistance $R_h = 10\,\Omega.$

2. Required power of heater

The heater must operate at $P_h = 62.5\,\text{W}.$

Since its resistance remains $10\,\Omega$, $P_h = \frac{V_h^2}{R_h}$ so $62.5 = \frac{V_h^2}{10}$ $V_h^2 = 625$ $V_h = 25\,\text{V}.$

Thus the heater should get $25\,\text{V}.$

Then current through heater is $I_h = \frac{V_h}{R_h} = \frac{25}{10} = 2.5\,\text{A}.$

3. Interpret the combination

The standard arrangement for this question is that the heater is connected in parallel with resistance $R$, and this parallel combination is in series with the $10\,\Omega$ resistor across the $100\,\text{V}$ supply.

Since the heater gets $25\,\text{V}$, the parallel branch has voltage $25\,\text{V}.$

Hence the series $10\,\Omega$ resistor must drop $100 - 25 = 75\,\text{V}.$

So current through the series $10\,\Omega$ resistor is $I_{\text{total}} = \frac{75}{10} = 7.5\,\text{A}.$

4. Find current through resistance $R$

Total current entering the parallel combination is $7.5\,\text{A}$.

Heater current is $2.5\,\text{A}$, so current through $R$ is $I_R = 7.5 - 2.5 = 5\,\text{A}.$

Voltage across $R$ is also $25\,\text{V}$, therefore $R = \frac{V}{I} = \frac{25}{5} = 5\,\Omega.$

5. Final answer

$\boxed{5\,\Omega}$

This matches the stored correct answer.