(b) Here, $i = 4A;V = 20\ $ Volt; so,
$R = \frac{V}{I} = \frac{20}{4} = 5A.$ Since, voltmeter is connected in parallel with resistance $R,$ the effective resistance of this combination is 5 $\Omega$ only if the resistance R is greater than 5$\Omega$, since total resistance in parallel combination becomes less than individual resistance.