To measure the potential difference across the 100 resistor in the...

The voltmeter is connected in parallel with the $100\text{ }\Omega$ resistor.

The equivalent resistance of the voltmeter and the $100\text{ }\Omega$ resistor is:
$R_p = \dfrac{100 \times 400}{100 + 400} = \dfrac{40000}{500} = 80\text{ }\Omega$

This equivalent resistance $R_p$ is in series with the $200\text{ }\Omega$ resistor. The total resistance of the circuit is:
$R_{eq} = R_p + 200 = 80 + 200 = 280\text{ }\Omega$

The total current drawn from the battery is:
$I = \dfrac{V}{R_{eq}} = \dfrac{84}{280} = 0.3\text{ A}$

The reading of the voltmeter is the potential difference across the parallel combination:
$V_{reading} = I \times R_p = 0.3 \times 80 = 24\text{ V}$